Chapter 3 - "Stoichiometry" - Long Questions

"Stoichiometry" - Long Questions

Question 1 

Hydrogen reacts with oxygen to form water.

(a) Write the word equation for this reaction.

(b) Write the correct balanced symbol equation.

(c) Explain why the symbol equation must be balanced.

Answer

(a) Hydrogen + Oxygen → Water

(b)2H₂ + O₂ → 2H₂O

(c)The symbol equation must be balanced to obey the law of conservation of mass. This law states that atoms are not created or destroyed during a chemical reaction. In the balanced equation, there are four hydrogen atoms and two oxygen atoms on both sides of the equation, showing that the number of each type of atom is the same before and after the reaction.

Question 2 

Magnesium burns in oxygen to form magnesium oxide.

(a) Write the chemical formula of magnesium oxide.

(b) Write a balanced symbol equation for the reaction.

(c) State the physical state of magnesium oxide at room temperature.

Answer

(a) MgO

(b) 2Mg + O₂ → 2MgO

(c) Magnesium oxide is a solid at room temperature.

Question 3 

The diagram shows atoms of two elements combined together.

(a) Deduce the chemical formula shown.

(b) Name the type of substance formed.

(c) Explain how the formula was deduced from the diagram.

Answer

(a) The chemical formula is AB₂.

(b) The substance formed is a compound.

(c) The diagram shows one atom of element A combined with two atoms of element B. Chemical formulae represent the simplest whole number ratio of atoms present, so the formula is shown as AB₂.

Question 4 

Sodium reacts with chlorine to form sodium chloride.

(a) State the chemical formula of sodium chloride.

(b) Identify the type of bonding present.

(c) Explain how charge balance is achieved in sodium chloride.

Answer

(a) NaCl

(b) Ionic bonding

(c) Sodium forms a positive ion, Na⁺, by losing one electron. Chlorine forms a negative ion, Cl⁻, by gaining one electron. The charges balance because one positive charge cancels one negative charge, resulting in an overall neutral compound.

Question 5 

Explain what is meant by the term law of conservation of mass and describe how it applies to chemical reactions.

Answer

The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. This means the total mass of the reactants is equal to the total mass of the products. In chemical reactions, atoms are rearranged but not lost or gained. Balanced chemical equations show this by having the same number of each type of atom on both sides of the equation.

Question 6 

Calcium reacts with oxygen to form calcium oxide.

(a) Write the formula of calcium oxide.

(b) Write the balanced symbol equation.

(c) State the state symbols for all substances.

Answer

(a) CaO

(b) 2Ca + O₂ → 2CaO

(c) 2Ca(s) + O₂(g) → 2CaO(s)

Question 7 

Explain the difference between a word equation and a symbol equation, using an example.

Answer

A word equation uses the names of substances to describe a chemical reaction, making it simple to understand. For example,

carbon + oxygen → carbon dioxide

A symbol equation uses chemical formulae and shows the exact ratios of reactants and products. For example,

C + O₂ → CO₂

Symbol equations provide more detailed information and must be balanced to show conservation of mass.

Question 8 

An ionic compound contains aluminium ions (Al³⁺) and oxide ions (O²⁻).

(a) Deduce the formula of the compound.

(b) Explain how the charges are balanced.

Answer

(a) Al₂O₃

(b)

Two aluminium ions produce a total positive charge of +6, while three oxide ions produce a total negative charge of −6. These charges balance, giving an overall neutral compound.

Question 9 

What information do state symbols provide in chemical equations?

Name and explain the meaning of all four state symbols.

Answer

State symbols show the physical state of substances in a chemical reaction.

(s) means solid,

(l) means liquid,

(g) means gas,

(aq) means aqueous, which indicates the substance is dissolved in water.

State symbols help explain how reactions occur and under what conditions they take place.

Question 10 

Explain why chemical formulae are essential in stoichiometric calculations.

Answer

Chemical formulae show the types and numbers of atoms in a substance. Stoichiometric calculations rely on these ratios to determine how much reactant is needed or how much product will form. Without correct formulae, it would be impossible to balance equations, apply mole ratios, or predict quantities accurately. Chemical formulae therefore form the foundation of all quantitative chemistry.

Question 11 

(a) Define the term element.

(b) State the chemical symbols for sodium, iron, copper, and chlorine.

(c) Explain why correct capitalization of symbols is important.

Answer

(a) An element is a pure substance made of only one type of atom.

(b)

• Sodium: Na
• Iron: Fe
• Copper: Cu
• Chlorine: Cl

(c) Correct capitalization is important because chemical symbols are case-sensitive. A capital letter followed by a lowercase letter represents a specific element. For example, Co represents cobalt, while CO represents carbon monoxide, which is a compound. Incorrect capitalization can completely change the meaning.

Question 12 

Some elements exist naturally as single atoms.

(a) Name two metals that exist as single atoms.

(b) Name two noble gases that exist as single atoms.

(c) State the formula used to represent these elements.

Answer

(a) Magnesium and iron

(b) Helium and neon

(c) These elements are represented using only their chemical symbols, such as Mg, Fe, He, and Ne, with no subscripts.

Question 13

Explain why most metals are written as single atoms rather than molecules.

Answer

Most metals are written as single atoms because they form giant metallic structures rather than small molecules. In these structures, positive metal ions are arranged in a lattice and surrounded by delocalized electrons. Since metals do not exist as individual molecules, their chemical formulae are written using only their symbols.

Question 14 

Hydrogen, oxygen, and nitrogen exist naturally as molecules.

(a) State the formula of each of these elements.

(b) Describe what the subscript 2 represents.

(c) Name the type of elements that exist in this form.

Answer

(a)

• Hydrogen: H₂
• Oxygen: O₂
• Nitrogen: N₂

(b) The subscript 2 shows that two atoms of the same element are chemically bonded together in one molecule.

(c) These elements are known as diatomic elements.

Question 15

Chlorine exists as Cl₂ and not Cl.

(a) State why chlorine is written as Cl₂.

(b) State one error that would occur if chlorine were written as Cl in equations.

Answer

(a) Chlorine exists as diatomic molecules because two chlorine atoms bond together to form a stable molecule.

(b) Writing chlorine as Cl would show only one atom instead of two. This would lead to incorrect balancing of equations and wrong stoichiometric calculations.

Question 16 

The list below contains several elements:

oxygen, magnesium, bromine, neon

(a) Write the correct chemical formula for each element.

(b) Identify which exist as single atoms and which exist as molecules.

Answer

(a)

• Oxygen: O₂
• Magnesium: Mg
• Bromine: Br₂
• Neon: Ne

(b)

• Single atoms: magnesium and neon
• Molecules: oxygen and bromine

Question 17 

Explain why noble gases exist as single atoms.

Answer

Noble gases exist as single atoms because they already have full outer electron shells. This makes them very stable and unreactive. As a result, noble gas atoms do not need to bond with other atoms, so they exist as individual atoms rather than molecules.

Question 18 

A student writes the formula for oxygen gas as O.

(a) State why this is incorrect.

(b) Give the correct formula.

(c) Explain how this mistake would affect chemical calculations.

Answer

(a) This is incorrect because oxygen exists naturally as a diatomic molecule.

(b) The correct formula is O₂.

(c) Using O instead of O₂ would give the wrong number of atoms in equations, leading to incorrect balancing and incorrect calculation of reacting masses or volumes.

Question 19 

State the formulae of the seven common diatomic elements that students must know for IGCSE Chemistry.

Answer

The seven common diatomic elements are:

H₂, N₂, O₂, F₂, Cl₂, Br₂, and I₂.

These elements always exist as molecules containing two atoms when they are in their elemental form.

Question 20 

Explain why correct elemental formulae are essential in chemical equations.

Answer

Correct elemental formulae are essential because chemical equations represent the actual particles involved in reactions. Writing incorrect formulae changes the number of atoms present, which leads to incorrect balancing of equations and incorrect stoichiometric calculations. Accurate formulae ensure that equations obey the law of conservation of mass and that chemical reactions are represented correctly.

Question 21 

(a) Define the term compound.

(b) State two ways in which compounds differ from elements.

Answer

(a) A compound is a pure substance formed when two or more different elements are chemically combined in fixed proportions.

(b) Compounds contain more than one type of atom, whereas elements contain only one type of atom.

Compounds have properties that are different from the elements that form them because their atoms are chemically bonded together.

Question 22

Water is a covalent compound.

(a) State the chemical formula of water.

(b) Explain what information the formula shows.

(c) State the type of bonding present in water.

Answer

(a) H₂O

(b) The formula shows that each molecule of water contains two hydrogen atoms and one oxygen atom.

(c) Covalent bonding

Question 23 

Carbon monoxide and carbon dioxide are two different compounds.

(a) State the formula of carbon monoxide.

(b) State the formula of carbon dioxide.

(c) Explain why these two compounds have different properties.

Answer

(a) CO

(b) CO₂

(c) The two compounds have different numbers of oxygen atoms bonded to carbon. This difference in composition leads to different structures and different chemical and physical properties, even though both compounds contain the same elements.

Question 24 

Methane is an important covalent compound.

(a) State the chemical formula of methane.

(b) Name the elements present in methane.

(c) State how many atoms are present in one molecule of methane.

Answer

(a) CH₄

(b) Carbon and hydrogen

(c) One molecule of methane contains five atoms in total, one carbon atom and four hydrogen atoms.

Question 25 

Sodium chloride is an ionic compound.

(a) State the chemical formula of sodium chloride.

(b) Identify the ions present.

(c) Explain why the formula contains no subscripts.

Answer

(a) NaCl

(b) Sodium ions, Na⁺, and chloride ions, Cl⁻

(c) The charges on the ions balance in a 1:1 ratio. One positive charge from sodium balances one negative charge from chloride, so no subscripts are required.

Question 26 

Calcium chloride has the formula CaCl₂.

(a) State the charges on the calcium and chloride ions.

(b) Explain why the formula contains a subscript 2.

Answer

(a)

• Calcium ion: Ca²⁺
• Chloride ion: Cl⁻

(b) One calcium ion has a charge of +2, while each chloride ion has a charge of −1. Two chloride ions are needed to balance the charge of one calcium ion, giving the formula CaCl₂.

Question 27 

Aluminium oxide has the formula Al₂O₃.

(a) State the charges on the aluminium and oxide ions.

(b) Show how the charges balance to form a neutral compound.

Answer

(a)

• Aluminium ion: Al³⁺
• Oxide ion: O²⁻

(b)

• Two aluminium ions give a total positive charge of +6.
• Three oxide ions give a total negative charge of −6.
• These charges balance, resulting in an overall neutral compound.

Question 28 

Explain the role of subscripts in chemical formulae, using H₂SO₄ as an example.

Answer

Subscripts show how many atoms of each element are present in a compound. In H₂SO₄, the subscript 2 shows there are two hydrogen atoms, and the subscript 4 shows there are four oxygen atoms. Sulfur has no subscript, which means there is one sulfur atom. Subscripts apply only to the element immediately before them.

Question 29 

Explain what is meant by fixed ratios in compounds.

Answer

Fixed ratios mean that the elements in a compound always combine in the same proportion. For example, water is always H₂O and carbon dioxide is always CO₂. These ratios do not change because atoms bond in specific ways to achieve stable electron arrangements. This is described by the law of definite proportions.

Question 30 

Explain why stating the correct chemical formula of a compound is essential in chemistry.

Answer

Correct chemical formulae are essential because they show the exact composition of a compound. Formulae are used to write balanced equations, perform calculations, and predict the outcomes of reactions. An incorrect formula leads to incorrect equations and incorrect calculations, which affects the accuracy of chemical work and understanding.

Question 31 

(a) Define the term molecular formula.

(b) State one type of compound for which molecular formulae are used.

Answer

(a) The molecular formula of a compound shows the type and number of each atom present in one molecule of the compound.

(b) Molecular formulae are used for covalent compounds.

Question 32 

Explain the difference between a molecule and a compound.

Answer

A molecule is a group of atoms held together by covalent bonds. These atoms may be the same element or different elements.

A compound is a pure substance formed when two or more different elements are chemically combined in fixed proportions.

All molecular compounds are molecules, but not all molecules are compounds.

Question 33 

Water has the molecular formula H₂O.

(a) Name the elements present.

(b) State the number of atoms of each element in one molecule.

(c) Explain what the absence of a subscript means.

Answer

(a)Hydrogen and oxygen

(b)Two hydrogen atoms and one oxygen atom

(c) The absence of a subscript means there is only one atom of that element present in the molecule.

Question 34 

Carbon monoxide and carbon dioxide have the molecular formulae CO and CO₂.

(a) State one similarity between these compounds.

(b) State one difference in their molecular formulae.

(c) Explain why their properties are different.

Answer

(a)Both compounds contain carbon and oxygen.

(b) Carbon monoxide contains one oxygen atom, while carbon dioxide contains two oxygen atoms.

(c) The different number of oxygen atoms changes the structure and bonding in the molecules, resulting in very different chemical and physical properties.

Question 35 

Ammonia has the molecular formula NH₃.

(a) State the number of atoms present in one molecule of ammonia.

(b) Name the elements present.

(c) State the type of bonding in ammonia.

Answer

(a) Four atoms in total

(b) Nitrogen and hydrogen

(c) Covalent bonding

Question 36 

Explain the role of chemical symbols and subscripts in molecular formulae.

Answer

Chemical symbols show which elements are present in a molecule. Subscripts show how many atoms of each element are present in one molecule. A subscript applies only to the element immediately before it. If no subscript is written, it means there is one atom of that element. Together, symbols and subscripts allow the exact atomic composition of a molecule to be shown clearly.

Question 37 

Methane has the molecular formula CH₄.

(a) State how many hydrogen atoms are bonded to carbon.

(b) Explain why methane always has this formula.

Answer

(a) Four hydrogen atoms are bonded to one carbon atom.

(b) Methane has a fixed composition because carbon forms four covalent bonds to achieve a stable electron arrangement. This results in the same ratio of atoms in every methane molecule.

Question 38 

Explain the difference between a molecular formula and an empirical formula, using glucose as an example.

Answer

The molecular formula shows the actual number of atoms in one molecule, while the empirical formula shows the simplest whole-number ratio of atoms.

For glucose, the molecular formula is C₆H₁₂O₆, while the empirical formula is CH₂O. Both describe the same compound, but the molecular formula gives more detailed information.

Question 39 

Explain why molecular formulae show that compounds have a fixed composition.

Answer

Molecular formulae show that the atoms in a compound are combined in fixed ratios. Each molecule of a compound contains the same number and type of atoms. For example, water is always H₂O and carbon dioxide is always CO₂. This fixed composition results from the way atoms bond to achieve stable electron arrangements and explains why compounds have consistent properties.

Question 40 

Explain how molecular formulae are used in chemical reactions and equations.

Answer

Molecular formulae are used to write correct chemical equations. They allow chemists to track the number of atoms of each element before and after a reaction. Using correct molecular formulae ensures equations can be balanced and that the law of conservation of mass is obeyed. Molecular formulae also allow accurate calculations of reacting quantities and products formed.

Question 41 

A molecule made of one carbon atom bonded to four hydrogen atoms.

(a) State the symbols of the elements present.

(b) State the number of atoms of each element.

(c) Deduce the chemical formula of the compound.

Answer

(a) Carbon (C) and hydrogen (H)

(b) One carbon atom and four hydrogen atoms

(c) The chemical formula is CH₄

Question 42 

There are two hydrogen atoms for every one oxygen atom.

(a) State the ratio of hydrogen atoms to oxygen atoms.

(b) Deduce the chemical formula of the compound.

(c) Explain why subscripts are used in the formula.

Answer

(a) The ratio of hydrogen to oxygen atoms is 2:1

(b) The chemical formula is H₂O

(c) Subscripts show the number of atoms of each element present in one molecule. The subscript 2 shows there are two hydrogen atoms bonded to one oxygen atom.

Question 43 

A diagram shows four hydrogen atoms and two oxygen atoms arranged as two identical molecules.

(a) State the total number of hydrogen atoms shown.

(b) State the total number of oxygen atoms shown.

(c) Deduce the correct chemical formula of the compound and explain your answer.

Answer

(a) Four hydrogen atoms

(b) Two oxygen atoms

(c) The correct formula is H₂O. Chemical formulae are written using the simplest whole number ratio of atoms. The ratio of hydrogen to oxygen atoms is 2:1, even though more atoms are shown in total.

Question 44 

A diagram shows one nitrogen atom at the centre with three hydrogen atoms bonded to it.

(a) Name the central atom.

(b) State the number of surrounding atoms.

(c) Deduce the chemical formula of the compound.

Answer

(a) Nitrogen

(b) Three hydrogen atoms

(c) The chemical formula is NH₃

Question 45 

Explain what is meant by the term relative numbers of atoms, using a diagram that shows four hydrogen atoms and two oxygen atoms as an example.

Answer

Relative numbers of atoms refer to the ratio in which different atoms are present. In the example, four hydrogen atoms and two oxygen atoms give a ratio of 2:1. Chemical formulae use this simplest ratio, so the formula is H₂O, not H₄O₂. The total number of atoms does not affect the formula.

Question 46 

A diagram shows one carbon atom bonded to one oxygen atom.

(a) Deduce the chemical formula of the compound.

(b) Name another compound formed by carbon and oxygen with a different ratio.

(c) State how the diagrams help distinguish between the two compounds.

Answer

(a) The chemical formula is CO

(b) Carbon dioxide, CO₂

(c) The diagrams show different numbers of oxygen atoms bonded to carbon. Different ratios lead to different chemical formulae and different compounds.

Question 47 

Describe a step-by-step method for deducing the formula of a simple compound from a particle diagram.

Answer

• First, identify the different types of atoms using colour, size, or labels.
• Second, count the number of atoms of each type carefully.
• Third, write down the chemical symbols of the elements present.
• Finally, use subscripts to show the number of atoms of each element and reduce the numbers to the simplest whole number ratio.

Question 48 

A diagram represents a repeating pattern containing one magnesium atom and two chlorine atoms.

(a) Identify the elements present.

(b) State the ratio of magnesium atoms to chlorine atoms.

(c) Deduce the chemical formula of the compound.

Answer

(a) Magnesium and chlorine

(b) The ratio of magnesium to chlorine atoms is 1:2

(c) The chemical formula is MgCl₂

Question 49 

Explain why the shape or arrangement of atoms in a diagram does not affect the chemical formula of a compound.

Answer

The chemical formula depends only on the number of each type of atom present, not on how the atoms are arranged. Different shapes may represent the same compound, but as long as the relative numbers of atoms are the same, the chemical formula remains unchanged.

Question 50 

Explain why deducing chemical formulae from diagrams is important in stoichiometry.

Answer

Deducing formulae from diagrams helps students understand that chemical formulae are based on actual numbers of atoms. This skill allows correct equations to be written, ensures accurate calculations, and reinforces the idea that atoms combine in fixed ratios. It also prevents reliance on memorisation and improves understanding of chemical reactions.

Question 51 

Magnesium reacts with oxygen to form magnesium oxide.

(a) Write the word equation for this reaction.

(b) Write the balanced symbol equation.

Answer

(a) Magnesium + oxygen → magnesium oxide

(b) 2Mg + O₂ → 2MgO

Question 52 

Explain the difference between a word equation and a symbol equation.

Answer

A word equation uses the names of substances to describe a chemical reaction and helps show clearly which reactants form which products.

A symbol equation uses chemical symbols and formulae and shows the exact composition and ratios of substances involved.

Symbol equations must be balanced to obey the law of conservation of mass, while word equations do not show quantities.

Question 53 

Zinc reacts with hydrochloric acid to form zinc chloride and hydrogen.

(a) Write the word equation.

(b) Write the balanced symbol equation.

Answer

(a) Zinc + hydrochloric acid → zinc chloride + hydrogen

(b) Zn + 2HCl → ZnCl₂ + H₂

Question 54 

Explain what is meant by a balanced chemical equation and why equations must be balanced.

Answer

A balanced chemical equation is one in which the number of atoms of each element is the same on both sides of the equation.

Equations must be balanced to obey the law of conservation of mass, which states that atoms are not created or destroyed during a chemical reaction.

Balancing ensures that the equation correctly represents what happens at the particle level.

Question 55 

Hydrogen reacts with oxygen to form water.

(a) Write the unbalanced symbol equation.

(b) Balance the equation.

(c) State what the coefficients represent.

Answer

(a) H₂ + O₂ → H₂O

(b) 2H₂ + O₂ → 2H₂O

(c) The coefficients show the number of molecules taking part in the reaction. Two molecules of hydrogen react with one molecule of oxygen to form two molecules of water.

Question 56 

State the meanings of the following state symbols:

(s), (l), (g), (aq)

Answer

(s) means solid

(l) means liquid

(g) means gas

(aq) means aqueous, meaning dissolved in water

State symbols show the physical state of substances during a chemical reaction.

Question 57 

Magnesium reacts with hydrochloric acid in solution.

(a) Write the balanced symbol equation including state symbols.

Answer

(a) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Question 58 

Explain why subscripts must not be changed when balancing chemical equations.

Answer

Subscripts are part of the chemical formula and show the number of atoms in a substance. Changing a subscript changes the identity of the compound.

Balancing equations must be done by adding coefficients in front of formulae, not by changing subscripts, to ensure the substances involved remain the same.

Question 59 

Calcium carbonate decomposes on heating to form calcium oxide and carbon dioxide.

(a) Write the word equation.

(b) Write the balanced symbol equation including state symbols.

Answer

(a) Calcium carbonate → calcium oxide + carbon dioxide

(b) CaCO₃(s) → CaO(s) + CO₂(g)

Question 60 

Explain why state symbols are important in chemical equations.

Answer

State symbols show the physical state of each substance in a reaction. They help explain observations such as gas production, formation of solids, or substances remaining in solution.

State symbols are especially important in reactions involving solutions, gases, or precipitates, as they give a clearer and more complete description of the chemical change.

Question 61 

(a) Define the term empirical formula.

(b) State one way in which an empirical formula differs from a molecular formula.

Answer

(a) The empirical formula of a compound is the simplest whole number ratio of the different atoms or ions present in the compound.

(b) An empirical formula shows only the simplest ratio of atoms, while a molecular formula shows the actual number of atoms in one molecule.

Question 62 

A compound contains carbon and hydrogen atoms in the ratio 2:4.

(a) State the simplest whole number ratio of carbon to hydrogen.

(b) Deduce the empirical formula of the compound.

(c) Explain why the ratio must be simplified.

Answer

(a) The simplest ratio is 1:2.

(b) The empirical formula is CH₂.

(c)Empirical formulae must show the smallest whole number ratio of atoms. Simplifying the ratio ensures the formula represents the basic composition of the compound.

Question 63 

Explain why empirical formulae always use whole numbers.

Answer

Empirical formulae use whole numbers because atoms and ions are discrete particles. Fractions of atoms cannot exist in compounds. If a ratio contains fractions, the numbers are multiplied until a whole number ratio is obtained, ensuring the formula represents real particles.

Question 64 

Water has the molecular formula H₂O.

(a) State the ratio of hydrogen atoms to oxygen atoms.

(b) Deduce the empirical formula of water.

(c) Explain why the empirical and molecular formulae are the same.

Answer

(a) The ratio of hydrogen to oxygen atoms is 2:1.

(b) The empirical formula is H₂O.

(c) The ratio 2:1 cannot be simplified further, so the empirical formula is the same as the molecular formula.

Question 65 

Glucose has the molecular formula C₆H₁₂O₆.

(a) Write the ratio of carbon, hydrogen, and oxygen atoms.

(b) Deduce the empirical formula of glucose.

Answer

(a) The ratio of carbon to hydrogen to oxygen atoms is 6:12:6.

(b) Dividing each number by 6 gives the simplest ratio 1:2:1, so the empirical formula is CH₂O.

Question 66 

Explain why empirical formulae are especially important for ionic compounds.

Answer

Ionic compounds do not exist as individual molecules. They form giant lattices made of repeating positive and negative ions. The formula of an ionic compound shows the simplest ratio of ions present, which is an empirical formula. This ratio ensures that the total charge of the compound is neutral.

Question 67 

Magnesium chloride contains magnesium ions and chloride ions.

(a) State the charges on the magnesium and chloride ions.

(b) Deduce the empirical formula of magnesium chloride.

(c) Explain how the charges are balanced.

Answer

(a)

Magnesium ion: Mg²⁺

Chloride ion: Cl⁻

(b)The empirical formula is MgCl₂.

(c)One magnesium ion has a +2 charge, and each chloride ion has a −1 charge. Two chloride ions are needed to balance the charge of one magnesium ion.

Question 68 

A diagram shows two red atoms and four white atoms bonded together repeatedly.

(a) State the ratio of red atoms to white atoms.

(b) Deduce the empirical formula using symbols R and W.

(c) Explain why total numbers are not used directly.

Answer

(a)The ratio of red atoms to white atoms is 2:4.

(b)The simplest ratio is 1:2, so the empirical formula is RW₂.

(c)Empirical formulae use the simplest ratio of atoms, not the total number shown in the diagram.

Question 69 

Explain how the empirical formula of a compound shows fixed composition.

Answer

The empirical formula shows the simplest ratio in which atoms or ions combine in a compound. This ratio is always the same for a given compound, regardless of sample size. This reflects the law of definite proportions, which states that a compound always contains the same elements in the same proportions.

Question 70

Ethene has the molecular formula C₂H₄ and propene has the molecular formula C₃H₆.

(a) Deduce the empirical formula of both compounds.

(b) Explain why these compounds can have the same empirical formula but different properties.

Answer

(a)

Ethene: C₂H₄ simplifies to CH₂

Propene: C₃H₆ simplifies to CH₂

(b) Although both compounds have the same empirical formula, they have different molecular formulae and structures. This leads to different molecular sizes and different chemical and physical properties.

Question 71 

(a) Define the term ion.

(b) Distinguish between a cation and an anion.

(c) State why ions form ionic compounds.

Answer

(a) An ion is an atom or group of atoms that has gained or lost electrons and therefore carries an electric charge.

(b) A cation is a positively charged ion formed by losing electrons, while an anion is a negatively charged ion formed by gaining electrons.

(c) Ions form ionic compounds because oppositely charged ions attract each other by strong electrostatic forces, resulting in a stable structure.

Question 72 

Explain what the chemical formula of an ionic compound represents.

Answer

The chemical formula of an ionic compound represents the simplest whole number ratio of positive and negative ions present in the giant ionic lattice. It does not represent a molecule but shows the smallest repeating ratio that gives an overall neutral compound.

Question 73

A diagram shows equal numbers of sodium ions and chloride ions arranged in a lattice.

(a) Identify the ions present.

(b) State the ratio of ions.

(c) Deduce the chemical formula of the compound.

Answer

(a) Sodium ions, Na⁺, and chloride ions, Cl⁻

(b) The ratio of sodium ions to chloride ions is 1:1.

(c)  The chemical formula is NaCl.

Question 74 

A diagram shows one magnesium ion surrounded by two chloride ions.

(a) State the charges on the magnesium and chloride ions.

(b) State the ratio of magnesium ions to chloride ions.

(c) Deduce the chemical formula of the compound.

Answer

(a)

• Magnesium ion: Mg²⁺
• Chloride ion: Cl⁻

(b) The ratio is 1:2.

(c) The chemical formula is MgCl₂.

Question 75 

Explain how the charges on ions are used to deduce the formula of an ionic compound.

Answer

The charges on ions are used to ensure that the total positive charge equals the total negative charge. By finding the smallest number of each ion needed to balance the charges, the simplest whole number ratio is obtained. This ratio is then written as the chemical formula using subscripts.

Question 76 

Calcium forms Ca²⁺ ions and oxygen forms O²⁻ ions.

(a) State the charge on each ion.

(b) Explain how the charges are balanced.

(c) Deduce the chemical formula of the compound.

Answer

(a)

Calcium ion: Ca²⁺

Oxide ion: O²⁻

(b) One calcium ion has a +2 charge and one oxide ion has a −2 charge. These charges cancel in a 1:1 ratio.

(c) The chemical formula is CaO.

Question 77 

Aluminium forms Al³⁺ ions and oxygen forms O²⁻ ions.

(a) Explain why one aluminium ion cannot combine with one oxide ion.

(b) Deduce the simplest ratio of aluminium ions to oxide ions.

(c) Write the chemical formula of the compound.

Answer

(a) One aluminium ion has a +3 charge and one oxide ion has a −2 charge, so the charges do not cancel.

(b) Two aluminium ions give a total charge of +6, and three oxide ions give a total charge of −6.

(c) The chemical formula is Al₂O₃.

Question 78

Explain the criss-cross method for writing ionic formulae and state one limitation of this method.

Answer

The criss-cross method involves crossing the charge numbers of the ions to become subscripts in the formula. This helps find the correct ratio of ions quickly.

A limitation is that the resulting formula must always be simplified, and students must understand charge balance rather than using the method mechanically.

Question 79 

Ammonium ions, NH₄⁺, combine with sulfate ions, SO₄²⁻.

(a) State the charge on each ion.

(b) Deduce the ratio of ammonium ions to sulfate ions.

(c) Write the chemical formula of the compound.

Answer

(a)

• Ammonium ion: +1
• Sulfate ion: −2

(b) Two ammonium ions are needed for each sulfate ion.

(c) The chemical formula is (NH₄)₂SO₄.

Question 80 

Explain why the formula of an ionic compound is always an empirical formula.

Answer

Ionic compounds do not exist as individual molecules. They form giant lattices made of repeating ions. The formula shows only the simplest whole number ratio of ions present in the lattice, which by definition is an empirical formula. There is no molecular formula for ionic compounds.

Question 81 

(a) Define the term symbol equation.

(b) State one advantage of using a symbol equation instead of a word equation.

Answer

(a) A symbol equation is a chemical equation that represents a reaction using chemical symbols and formulae instead of words.

(b) A symbol equation shows the exact substances involved and the ratios in which they react, allowing the equation to be balanced and used for calculations.

Question 82 

Magnesium reacts with oxygen to form magnesium oxide.

(a) Write the unbalanced symbol equation.

(b) Balance the equation.

(c) State the law used to balance equations.

Answer

(a) Mg + O₂ → MgO

(b) 2Mg + O₂ → 2MgO

(c) The law of conservation of mass.

Question 83 

Explain why coefficients are used when balancing symbol equations instead of changing subscripts.

Answer

Coefficients are used because they change the number of reacting particles without changing the identity of the substances. Changing subscripts would alter the chemical formula and create a different substance, which would make the equation incorrect.

Question 84 

Zinc reacts with hydrochloric acid to form zinc chloride and hydrogen.

(a) Write the balanced symbol equation.

(b) Add the correct state symbols.

Answer

(a) Zn + 2HCl → ZnCl₂ + H₂

(b) Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

Question 85 

State the meaning of each of the following state symbols and give one example for each:

(s), (l), (g), (aq)

Answer

(s) means solid, for example Mg(s).

(l) means liquid, for example H₂O(l).

(g) means gas, for example CO₂(g).

(aq) means aqueous, meaning dissolved in water, for example NaCl(aq).

Question 86 

Copper reacts with oxygen to form copper(II) oxide.

(a) Write the word equation.

(b) Write the balanced symbol equation including state symbols.

Answer

(a) Copper + oxygen → copper(II) oxide

(b) 2Cu(s) + O₂(g) → 2CuO(s)

Question 87 

Explain why state symbols are important in symbol equations.

Answer

State symbols show the physical state of each substance in a reaction. They help explain observations such as gas production, solid formation, or substances remaining in solution. They also distinguish between substances with the same formula but different states, such as HCl(g) and HCl(aq).

Question 88 

Silver nitrate solution reacts with sodium chloride solution.

(a) Write the balanced symbol equation including state symbols.

(b) State the colour of the precipitate formed.

Answer

(a) AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

(b) A white precipitate of silver chloride is formed.

Question 89 

The equation below represents a reaction in aqueous solution:

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

(a) Write the full ionic equation.

(b) Identify the spectator ions.

(c) Write the ionic equation.

Answer

(a) Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)

(b) Sodium ions and nitrate ions.

(c) Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Question 90 

Hydrochloric acid reacts with sodium hydroxide.

(a) Write the balanced symbol equation including state symbols.

(b) Write the ionic equation for this reaction.

Answer

(a) HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

(b) H⁺(aq) + OH⁻(aq) → H₂O(l)

Question 91 

A student is told that magnesium reacts with dilute hydrochloric acid to produce hydrogen gas and a soluble salt.

(a) Identify the reactants and products.

(b) Write the balanced symbol equation including state symbols.

Answer

(a)

• Reactants: magnesium and hydrochloric acid
• Products: magnesium chloride and hydrogen

(b) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Question 92 

Copper is heated strongly in air. A black solid is formed.

(a) Identify the reactants.

(b) Name the product formed.

(c) Write the balanced symbol equation including state symbols.

Answer

(a) Copper and oxygen

(b) Copper(II) oxide

(c) 2Cu(s) + O₂(g) → 2CuO(s)

Question 93 

Calcium carbonate is heated and decomposes to form calcium oxide and carbon dioxide.

(a) State the type of reaction.

(b) Write the balanced symbol equation including state symbols.

Answer

(a) Thermal decomposition

(b) CaCO₃(s) → CaO(s) + CO₂(g)

Question 94 

A metal reacts with dilute sulfuric acid. A gas is released that burns with a squeaky pop.

(a) Identify the gas produced.

(b) State the general word equation for this type of reaction.

(c) Write a balanced symbol equation for zinc reacting with sulfuric acid, including state symbols.

Answer

(a) Hydrogen

(b) Metal + acid → salt + hydrogen

(c) Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

Question 95

Silver nitrate solution is mixed with sodium chloride solution. A white solid forms.

(a) Name the white solid formed.

(b) Write the balanced symbol equation including state symbols.

Answer

(a) Silver chloride

(b) AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Question 96 

Hydrochloric acid reacts with sodium hydroxide solution.

(a) Identify the type of reaction.

(b) Write the balanced symbol equation including state symbols.

Answer

(a) Neutralization

(b) HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Question 97 

A student observes bubbling when calcium reacts with water. Calcium hydroxide solution and a gas are formed.

(a) Identify the gas produced.

(b) Write the balanced symbol equation including state symbols.

Answer

(a) Hydrogen

(b) Ca(s) + 2H₂O(l) → Ca(OH)₂(aq) + H₂(g)

Question 98

Lead(II) nitrate solution reacts with potassium iodide solution to form a yellow precipitate.

(a) Name the precipitate formed.

(b) Write the balanced symbol equation including state symbols.

Answer

(a) Lead(II) iodide

(b) Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

Question 99 

A student heats zinc carbonate. Zinc oxide and a gas are produced.

(a) Identify the gas formed.

(b) Write the balanced symbol equation including state symbols.

Answer

(a) Carbon dioxide

(b) ZnCO₃(s) → ZnO(s) + CO₂(g)

Question 100

Explain why state symbols are essential when deducing symbol equations from experimental information.

Answer

State symbols show the physical state of substances during a reaction. They help interpret observations such as gas production, formation of a precipitate, or substances remaining dissolved in water. Including correct state symbols ensures the equation accurately represents what happens in the reaction and allows full marks to be awarded in examinations.

Question 101 

(a) Define the term relative atomic mass (Ar).

(b) Name the reference standard used for defining relative atomic mass.

Answer

(a) Relative atomic mass (Ar) is the average mass of the isotopes of an element compared to 1/12th of the mass of an atom of carbon-12.

(b) The reference standard is the carbon-12 isotope.

Question 102 

Explain why a relative scale is used to compare atomic masses instead of measuring the actual masses of atoms.

Answer

Atoms are extremely small and have masses around 10⁻²⁷ kilograms, which are too small to measure directly or use conveniently. A relative scale allows chemists to compare atomic masses using manageable numbers. This makes calculations easier and allows consistent comparison of elements.

Question 103 

Carbon-12 is used as the standard for relative atomic mass.

(a) State how many protons and neutrons carbon-12 has.

(b) State the value assigned to carbon-12 on the relative atomic mass scale.

(c) Explain what one atomic mass unit represents.

Answer

(a) Carbon-12 has 6 protons and 6 neutrons.

(b) Carbon-12 is assigned a relative atomic mass of exactly 12.

(c)One atomic mass unit is defined as 1/12th of the mass of a carbon-12 atom.

Question 104 

Explain why carbon-12 was chosen as the reference standard for relative atomic mass.

Answer

Carbon-12 was chosen because it is stable, non-radioactive, abundant in nature, and easy to work with experimentally. It also forms many compounds, making it suitable for consistent and accurate measurements worldwide.

Question 105 

(a) Define the term isotope.

(b) Explain how isotopes affect the relative atomic mass of an element.

Answer

(a) Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons.

(b) Because isotopes have different masses and occur in different abundances, the relative atomic mass is calculated as an average of the isotopes present. More abundant isotopes contribute more to the average value.

Question 106

Chlorine exists naturally as two isotopes, chlorine-35 and chlorine-37.

(a) State why the relative atomic mass of chlorine is not a whole number.

(b) Explain why chlorine-35 affects the average more than chlorine-37.

Answer

(a)The relative atomic mass of chlorine is not a whole number because it is an average of the masses of its isotopes.

(b)Chlorine-35 is more abundant than chlorine-37, so it contributes more to the average atomic mass, giving a value of approximately 35.5.

Question 107

Explain the meaning of the phrase “compared to 1/12th of the mass of a carbon-12 atom”.

Answer

This phrase means that carbon-12 is assigned a mass of exactly 12 units. One atomic mass unit is therefore 1/12th of this mass. The masses of all other atoms are compared to this unit to determine their relative atomic masses.

Question 108 

Explain why relative atomic mass values shown on the Periodic Table are often decimals.

Answer

Relative atomic mass values are decimals because they represent the average mass of all naturally occurring isotopes of an element. Since isotopes have different masses and abundances, the average is often not a whole number.

Question 109 

State two uses of relative atomic mass in chemistry and explain one of them.

Answer

Relative atomic mass is used to calculate relative molecular mass and to calculate reacting masses in chemical reactions.

For example, it is used to calculate relative molecular mass by adding the Ar values of all atoms in a molecule, which is essential for mole and mass calculations.

Question 110 

Describe two common misconceptions about relative atomic mass and correct them.

Answer

One misconception is that relative atomic mass is the mass of a single atom. This is incorrect because Ar is a relative value, not an actual mass.

Another misconception is that Ar values must be whole numbers. In reality, Ar values are often decimals because they are averages of isotopes with different abundances.

Question 111 

(a) Define the term relative molecular mass (Mr).

(b) State what information is needed to calculate Mr.

Answer

(a) Relative molecular mass (Mr) is the sum of the relative atomic masses (Ar) of all the atoms present in one molecule of a substance.

(b) The chemical formula of the substance and the relative atomic masses of the elements from the Periodic Table are needed.

Question 112 

Explain why relative molecular mass has no units.

Answer

Relative molecular mass has no units because it is a relative value. It compares the mass of a molecule to 1/12th of the mass of a carbon-12 atom. Since it is a ratio and not an absolute mass, no units such as grams or kilograms are used.

Question 113 

Water has the molecular formula H₂O.

(a) State the relative atomic masses of hydrogen and oxygen.

(b) Calculate the relative molecular mass of water.

Answer

(a) Hydrogen has an Ar of 1 and oxygen has an Ar of 16.

(b) Mr of water = (2 × 1) + (1 × 16) = 18.

Question 114 

Carbon dioxide has the molecular formula CO₂.

(a) Identify the elements present.

(b) Calculate the relative molecular mass of carbon dioxide.

(c) State one reason why carbon dioxide has a higher Mr than water.

Answer

(a) Carbon and oxygen.

(b) Mr of carbon dioxide = (1 × 12) + (2 × 16) = 44.

(c) Carbon dioxide contains heavier atoms and a greater total atomic mass than water.

Question 115 

Ammonia has the molecular formula NH₃.

(a) State the relative atomic masses of nitrogen and hydrogen.

(b) Calculate the relative molecular mass of ammonia.

Answer

(a) Nitrogen has an Ar of 14 and hydrogen has an Ar of 1.

(b) Mr of ammonia = (1 × 14) + (3 × 1) = 17.

Question 116 

Methane has the formula CH₄ and ethane has the formula C₂H₆.

(a) Calculate the relative molecular mass of methane.

(b) Calculate the relative molecular mass of ethane.

(c) Explain why ethane has a higher Mr than methane.

Answer

(a) Mr of methane = (1 × 12) + (4 × 1) = 16.

(b) Mr of ethane = (2 × 12) + (6 × 1) = 30.

(c) Ethane has more atoms and therefore a greater total atomic mass than methane.

Question 117 

Explain the difference between relative molecular mass and relative formula mass.

Answer

Relative molecular mass is used for substances that exist as molecules, usually covalent compounds.

Relative formula mass is used for substances that do not exist as individual molecules, such as ionic compounds.

Both are calculated by adding the Ar values of the atoms in the formula, but the terms reflect the nature of the substance.

Question 118 

Sodium chloride has the formula NaCl.

(a) State why relative formula mass is used instead of relative molecular mass.

(b) Calculate the relative formula mass of sodium chloride.

Answer

(a) Sodium chloride does not exist as individual molecules. It forms a giant ionic lattice, so relative formula mass is used.

(b) Mr of NaCl = (1 × 23) + (1 × 35.5) = 58.5.

Question 119 

Magnesium oxide has the formula MgO.

(a) State the relative atomic masses of magnesium and oxygen.

(b) Calculate the relative formula mass of magnesium oxide.

Answer

(a) Magnesium has an Ar of 24 and oxygen has an Ar of 16.

(b) Mr of MgO = (1 × 24) + (1 × 16) = 40.

Question 120 

Aluminium oxide has the formula Al₂O₃.

(a) Calculate the relative formula mass of aluminium oxide.

(b) State one use of relative formula mass in chemistry.

Answer

(a) Mr of Al₂O₃ = (2 × 27) + (3 × 16) = 54 + 48 = 102.

(b) Relative formula mass is used to calculate reacting masses and to convert between mass and amount of substance in chemical calculations.

Question 121 

Magnesium reacts with oxygen to form magnesium oxide.

(a) Write the balanced symbol equation.

(b) Calculate the mass of oxygen that reacts with 12 g of magnesium.

Answer

(a) 2Mg + O₂ → 2MgO

(b)

Relative masses:

2Mg = 2 × 24 = 48

O₂ = 2 × 16 = 32

Mass ratio Mg : O = 48 : 32 = 3 : 2

If 3 g of Mg react with 2 g of O,

12 g of Mg react with (12 ÷ 3) × 2 = 8 g of oxygen.

Question 122 

Hydrogen reacts with oxygen to form water.

(a) Write the balanced symbol equation.

(b) Calculate the mass of oxygen needed to react with 5 g of hydrogen.

Answer

(a) 2H₂ + O₂ → 2H₂O

(b)

Relative masses:

2H₂ = 4

O₂ = 32

Mass ratio H : O = 4 : 32 = 1 : 8

If 1 g of hydrogen reacts with 8 g of oxygen,

5 g of hydrogen react with 40 g of oxygen.

Question 123 

Carbon reacts with oxygen to form carbon dioxide.

(a) Write the balanced symbol equation.

(b) Calculate the mass of oxygen required to react with 6 g of carbon.

Answer

(a) C + O₂ → CO₂

(b)

Relative masses:

Carbon = 12

O₂ = 32

Mass ratio C : O = 12 : 32

For 6 g of carbon:

(6 ÷ 12) × 32 = 16 g of oxygen.

Question 124 

Zinc reacts with hydrochloric acid according to the equation:

Zn + 2HCl → ZnCl₂ + H₂

(a) Calculate the relative mass of zinc.

(b) Calculate the mass of hydrochloric acid needed to react with 13 g of zinc.

Answer

(a) Relative mass of zinc = 65

(b)

Relative mass of 2HCl = 2 × 36.5 = 73

Mass ratio Zn : HCl = 65 : 73

For 13 g of zinc:

(13 ÷ 65) × 73 = 14.6 g of hydrochloric acid.

Question 125

Calcium carbonate decomposes when heated.

CaCO₃ → CaO + CO₂

(a) Calculate the relative formula mass of calcium carbonate.

(b) Calculate the mass of carbon dioxide produced from 50 g of calcium carbonate.

Answer

(a)

Mr of CaCO₃ = 40 + 12 + (3 × 16) = 100

(b)

100 g CaCO₃ produce 44 g CO₂

50 g CaCO₃ produce (50 ÷ 100) × 44 = 22 g of carbon dioxide.

Question 126 

Iron reacts with sulfur to form iron sulfide.

(a) Write the balanced symbol equation.

(b) Calculate the mass of sulfur required to react with 56 g of iron.

Answer

(a) Fe + S → FeS

(b)

Relative masses:

Iron = 56

Sulfur = 32

Mass ratio Fe : S = 56 : 32

So 56 g of iron react with 32 g of sulfur.

Question 127 

Methane burns in oxygen to form carbon dioxide and water.

(a) Write the balanced symbol equation.

(b) Calculate the mass of oxygen needed to react with 8 g of methane.

Answer

(a) CH₄ + 2O₂ → CO₂ + 2H₂O

(b)

Relative masses:

CH₄ = 16

2O₂ = 64

Mass ratio CH₄ : O₂ = 16 : 64 = 1 : 4

8 g of methane react with 32 g of oxygen.

Question 128 

Magnesium reacts with hydrochloric acid.

(a) Write the balanced symbol equation.

(b) Calculate the mass of hydrogen produced when 24 g of magnesium reacts completely.

Answer

(a) Mg + 2HCl → MgCl₂ + H₂

(b)

Relative masses:

Magnesium = 24

Hydrogen = 2

So 24 g of magnesium produce 2 g of hydrogen.

Question 129 

Calcium reacts with oxygen to form calcium oxide.

(a) Write the balanced symbol equation.

(b) Calculate the mass of calcium oxide formed from 20 g of calcium.

Answer

(a) 2Ca + O₂ → 2CaO

(b)

Relative masses:

2Ca = 2 × 40 = 80

2CaO = 2 × 56 = 112

So 80 g of calcium produce 112 g of calcium oxide

20 g of calcium produce (20 ÷ 80) × 112 = 28 g of calcium oxide.

Question 130 

Explain why reacting mass calculations at Core level do not require the mole concept.

Answer

At Core level, reacting mass calculations use balanced equations and relative masses to find fixed mass ratios. These ratios allow simple proportion calculations to be carried out directly. Because the particle ratios are already shown in the balanced equation, the mole concept is not needed to calculate reacting masses accurately.

Question 131 

(a) Define the term concentration of a solution.

(b) Explain the difference between a concentrated solution and a dilute solution.

Answer

(a) Concentration is a measure of the amount of solute dissolved in a fixed volume of solution.

(b) A concentrated solution contains a large amount of solute in a given volume, while a dilute solution contains a small amount of solute in the same volume of solution.

Question 132 

Explain why concentration is important in chemistry.

Answer

Concentration is important because it affects the speed of chemical reactions, the amount of product formed, and the accuracy of experimental results such as titrations. It also helps ensure solutions are used safely and effectively in laboratories and industry.

Question 133 

(a) State two units used to measure concentration in IGCSE Chemistry.

(b) Explain what each unit represents.

Answer

(a) g / dm³ and mol / dm³

(b) g / dm³ represents the mass of solute in grams dissolved in one cubic decimetre of solution.

mol / dm³ represents the amount of solute in moles dissolved in one cubic decimetre of solution.

Question 134 

(a) State the relationship between dm³, litres, and cm³.

(b) Explain why dm³ is commonly used in concentration calculations.

Answer

(a) 1 dm³ = 1 litre = 1000 cm³

(b) The unit dm³ is commonly used because it represents a convenient laboratory volume and allows concentration to be expressed per litre of solution.

Question 135 

A student dissolves 10 g of sodium chloride in water and makes the solution up to 1 dm³.

(a) Calculate the concentration in g / dm³.

(b) Explain what this value means.

Answer

(a) Concentration = 10 g ÷ 1 dm³ = 10 g / dm³

(b) This means that every cubic decimetre of the solution contains 10 grams of sodium chloride.

Question 136 

A solution contains 25 g of sugar dissolved in 2 dm³ of solution.

(a) Calculate the concentration in g / dm³.

(b) State whether this solution is more concentrated or less concentrated than a 20 g / dm³ solution.

Answer

(a) Concentration = 25 ÷ 2 = 12.5 g / dm³

(b) This solution is less concentrated than a 20 g / dm³ solution.

Question 137 

Explain two advantages of using g / dm³ as a unit of concentration.

Answer

One advantage is that g / dm³ is simple to use because mass and volume can be measured directly using a balance and a measuring cylinder.

Another advantage is that it does not require knowledge of atomic or molecular masses, making it suitable for simple laboratory and real-life applications.

Question 138 

(a) Define concentration measured in mol / dm³.

(b) Explain why mol / dm³ is often preferred in chemical reactions.

Answer

(a) Concentration in mol / dm³ is the amount of solute, in moles, dissolved in one cubic decimetre of solution.

(b) mol / dm³ is preferred because chemical reactions depend on the number of particles present. Using moles ensures accurate comparisons and calculations in reactions.

Question 139 

A sodium chloride solution has a concentration of 58.5 g / dm³.

(a) State the relative formula mass of sodium chloride.

(b) Deduce the concentration of the solution in mol / dm³.

Answer

(a) The relative formula mass of sodium chloride is 58.5.

(b) Since 58.5 g corresponds to 1 mole, the concentration is 1 mol / dm³.

Question 140 

Explain what happens to the concentration of a solution during dilution.

Answer

During dilution, solvent is added to the solution, increasing the volume while the amount of solute remains the same. As a result, the concentration decreases because the same amount of solute is spread over a larger volume.

Question 141 

(a) Define the term mole.

(b) State the value of the Avogadro constant.

Answer

(a) The mole is the amount of substance that contains 6.02 × 10²³ particles.

(b) The Avogadro constant is 6.02 × 10²³ particles per mole.

Question 142 

Explain what is meant by the term amount of substance in chemistry.

Answer

Amount of substance refers to the number of particles present in a sample. These particles may be atoms, molecules, ions, or formula units. It is measured in moles and does not refer to mass or volume.

Question 143 

Explain why chemists use the mole instead of counting particles directly.

Answer

Particles such as atoms and molecules are extremely small, so even a small mass contains an enormous number of particles. Counting them individually is impossible. The mole provides a practical way to count particles indirectly using measurable mass.

Question 144 

One mole of carbon atoms and one mole of magnesium atoms are compared.

(a) State how many atoms are present in each sample.

(b) Explain why the masses of the two samples are different.

Answer

(a) Each sample contains 6.02 × 10²³ atoms.

(b) The masses are different because carbon and magnesium atoms have different relative atomic masses. Magnesium atoms are heavier than carbon atoms.

Question 145

Explain what is meant by one mole of molecules, using water as an example.

Answer

One mole of molecules means 6.02 × 10²³ molecules of a substance. For water, one mole contains 6.02 × 10²³ water molecules. Each molecule contains two hydrogen atoms and one oxygen atom, but the mole counts molecules, not individual atoms.

Question 146 

Explain what is meant by one mole of formula units in ionic compounds.

Answer

Ionic compounds do not exist as molecules but as giant lattices. One mole of an ionic compound contains 6.02 × 10²³ formula units. For example, one mole of sodium chloride contains 6.02 × 10²³ NaCl formula units.

Question 147 

Describe the link between the mole and relative atomic or molecular mass.

Answer

One mole of a substance has a mass in grams equal to its relative atomic mass or relative molecular mass. For example, one mole of oxygen atoms has a mass of 16 g, and one mole of water molecules has a mass of 18 g.

Question 148 

Explain why different substances have different masses per mole, even though one mole always contains the same number of particles.

Answer

One mole of any substance contains the same number of particles, but different particles have different masses. Heavier atoms or molecules result in a greater mass per mole than lighter ones.

Question 149 

Explain why the mole can be described as a counting unit, similar to a dozen.

Answer

A dozen always represents 12 items, regardless of what the items are. Similarly, a mole always represents 6.02 × 10²³ particles, regardless of whether they are atoms, molecules, or ions. Both are counting units.

Question 150 

Use the equation below to explain how the mole is used in chemical reactions.

2H₂ + O₂ → 2H₂O

Answer

The equation shows that two moles of hydrogen molecules react with one mole of oxygen molecules to form two moles of water molecules. The mole allows the particle ratios in the equation to be applied to measurable amounts of substances in reactions.

Question 151 

(a) Write the formula that links mass, amount of substance, and molar mass.

(b) Explain what each term in the formula represents.

Answer

(a) amount of substance (mol) = mass (g) ÷ molar mass (g/mol)

(b)

• Mass is the measured quantity in grams.
• Amount of substance is measured in moles and represents the number of particles.
• Molar mass is the mass of one mole of a substance and depends on the type of particles present.

Question 152 

Magnesium has a molar mass of 24 g/mol.

(a) Calculate the amount of substance in moles present in 48 g of magnesium.

(b) State what this value means in terms of particles.

Answer

(a) Amount of substance = 48 ÷ 24 = 2 mol

(b) This means the sample contains 2 moles of magnesium atoms, which is twice the Avogadro number of atoms.

Question 153 

A sample of sodium chloride has a mass of 58.5 g.

(a) State the molar mass of sodium chloride.

(b) Calculate the amount of substance in moles.

Answer

(a) The molar mass of sodium chloride is 58.5 g/mol.

(b) Amount of substance = 58.5 ÷ 58.5 = 1 mol

Question 154 

Water has a molar mass of 18 g/mol.

(a) Calculate the number of moles in 9 g of water.

(b) Explain why this value is less than 1 mol.

Answer

(a) Amount of substance = 9 ÷ 18 = 0.5 mol

(b) The mass is less than the molar mass, so it contains fewer particles than one full mole.

Question 155 

(a) Rearrange the mole equation to calculate mass.

(b) Calculate the mass of 2 moles of carbon dioxide.

Answer

(a) mass (g) = amount of substance (mol) × molar mass (g/mol)

(b) Molar mass of CO₂ = 44 g/mol

Mass = 2 × 44 = 88 g

Question 156

A sample contains 0.25 mol of magnesium oxide (MgO).

(a) Calculate the molar mass of magnesium oxide.

(b) Calculate the mass of the sample.

Answer

(a) Molar mass of MgO = 24 + 16 = 40 g/mol

(b) Mass = 0.25 × 40 = 10 g

Question 157 

A substance has a mass of 20 g and an amount of substance of 0.5 mol.

(a) Calculate the molar mass of the substance.

(b) Explain how this value can help identify the substance.

Answer

(a) Molar mass = 20 ÷ 0.5 = 40 g/mol

(b) The calculated molar mass can be compared with known relative atomic or molecular masses to identify the substance.

Question 158 

Explain the relationship between molar mass and relative atomic or molecular mass.

Answer

Numerically, molar mass in g/mol is equal to relative atomic mass or relative molecular or formula mass. The difference is that molar mass has units, while relative mass does not. This allows experimental masses to be linked to atomic and molecular structure.

Question 159 

A sample contains 0.5 mol of oxygen molecules.

(a) State the Avogadro constant.

(b) Calculate the number of oxygen molecules in the sample.

Answer

(a) The Avogadro constant is 6.02 × 10²³ particles per mole.

(b) Number of molecules = 0.5 × 6.02 × 10²³ = 3.01 × 10²³ molecules

Question 160

Describe the two-step method used to calculate the number of particles from a given mass of a substance.

Answer

First, the mass is converted into moles using the formula amount of substance = mass ÷ molar mass.

Second, the number of particles is calculated by multiplying the number of moles by the Avogadro constant.

This method links measurable mass to the number of particles present.

Question 161 

(a) State the molar gas volume at room temperature and pressure (r.t.p.).

(b) Explain what is meant by r.t.p.

Answer

(a) The molar gas volume at r.t.p. is 24 dm³ per mole.

(b) Room temperature and pressure refers to conditions of about 20 °C and 1 atmosphere (approximately 101 kPa), under which gases show predictable behavior.

Question 162 

Explain why all gases have the same molar volume at r.t.p.

Answer

Gas particles are very far apart and the volume of the particles themselves is negligible. At the same temperature and pressure, the volume of a gas depends only on the number of particles present. Since one mole of any gas contains the same number of particles, all gases occupy the same volume of 24 dm³ at r.t.p.

Question 163 

State the relationship between dm³ and cm³, and convert 24 dm³ into cm³.

Answer

1 dm³ = 1000 cm³

24 dm³ = 24 × 1000 = 24 000 cm³

Question 164 

Calculate the volume occupied by 2.5 mol of oxygen gas at r.t.p.

Answer

Volume of gas = amount of substance × 24

Volume = 2.5 × 24 = 60 dm³

Question 165 

A gas sample occupies a volume of 6 dm³ at r.t.p.

(a) Calculate the amount of substance in moles.

(b) State what this value represents.

Answer

(a) Amount of substance = volume ÷ 24 = 6 ÷ 24 = 0.25 mol

(b) This means the sample contains one quarter of a mole of gas particles.

Question 166 

Hydrogen gas reacts with oxygen according to the equation:

2H₂(g) + O₂(g) → 2H₂O(g)

(a) State the volume of hydrogen that reacts with 24 dm³ of oxygen at r.t.p.

(b) State the volume of water vapour formed.

Answer

(a)

The mole ratio H₂ : O₂ is 2 : 1.

So the volume ratio is also 2 : 1.

24 dm³ of oxygen reacts with 48 dm³ of hydrogen.

(b) 2 moles of water vapour are formed, which occupy 48 dm³ at r.t.p.

Question 167 

Calcium carbonate reacts with hydrochloric acid to produce carbon dioxide:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

(a) State the mole ratio of CaCO₃ to CO₂.

(b) Calculate the volume of carbon dioxide produced from 0.5 mol of calcium carbonate.

Answer

(a) The mole ratio CaCO₃ : CO₂ is 1 : 1.

(b) 0.5 mol of CO₂ occupies: 0.5 × 24 = 12 dm³

Question 168 

Explain how the molar gas volume can be used to identify the limiting reactant in a reaction that produces a gas.

Answer

The balanced equation shows the mole ratio of reactants and products. By calculating the number of moles of each reactant, the limiting reactant can be identified as the one that produces the smaller amount of gas. The molar gas volume is then used to calculate the volume of gas formed from the limiting reactant.

Question 169 

Two gas samples each occupy 24 dm³ at r.t.p., one is hydrogen and the other is carbon dioxide.

(a) Compare the number of molecules in each sample.

(b) Explain why their masses are different.

Answer

(a) Both samples contain the same number of molecules, equal to one mole.

(b) Carbon dioxide molecules are heavier than hydrogen molecules, so even though the number of particles is the same, the mass of carbon dioxide is much greater.

Question 170 

Describe two advantages of using gas volume calculations at r.t.p. instead of mass measurements.

Answer

One advantage is that gas volume can be measured directly using apparatus such as gas syringes.

Another advantage is that the molar gas volume allows quick conversion between moles and volume without needing the molar mass of the gas.

Question 171 

Magnesium reacts with oxygen according to the equation:

2Mg(s) + O₂(g) → 2MgO(s)

(a) Calculate the mass of oxygen needed to react completely with 12 g of magnesium.

(b) State which law of chemistry this calculation is based on.

Answer

(a)

Relative masses:

2Mg = 2 × 24 = 48

O₂ = 2 × 16 = 32

Mass ratio Mg : O = 48 : 32 = 3 : 2

For 12 g of magnesium:

(12 ÷ 3) × 2 = 8 g of oxygen

(b)This calculation is based on the law of conservation of mass.

Question 172 

Zinc reacts with hydrochloric acid:

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

Given 6.5 g of zinc and 7.3 g of hydrochloric acid:

(a) Calculate the moles of each reactant.

(b) Identify the limiting reactant.

Answer

(a)

Moles of Zn = 6.5 ÷ 65 = 0.10 mol

Moles of HCl = 7.3 ÷ 36.5 = 0.20 mol

(b)

The equation requires 2 mol of HCl per 1 mol of Zn.

The reactants are in the exact ratio, so neither reactant is limiting.

Question 173 

Carbon reacts with oxygen to form carbon dioxide:

C(s) + O₂(g) → CO₂(g)

(a) Calculate the mass of oxygen needed to react with 6 g of carbon.

(b) Calculate the volume of carbon dioxide formed at r.t.p.

Answer

(a)

Relative masses:

C = 12

O₂ = 32

For 6 g of carbon:

(6 ÷ 12) × 32 = 16 g of oxygen

(b)

Moles of CO₂ = 6 ÷ 12 = 0.5 mol

Volume = 0.5 × 24 = 12 dm³

Question 174 

Calcium carbonate decomposes when heated:

CaCO₃(s) → CaO(s) + CO₂(g)

(a) Calculate the mass of carbon dioxide formed from 50 g of calcium carbonate.

(b) Calculate the volume of carbon dioxide produced at r.t.p.

Answer

(a)

Mr of CaCO₃ = 100

CO₂ formed from 100 g = 44 g

From 50 g:

(50 ÷ 100) × 44 = 22 g of CO₂

(b)

Moles of CO₂ = 22 ÷ 44 = 0.5 mol

Volume = 0.5 × 24 = 12 dm³

Question 175 

Hydrogen reacts with oxygen:

2H₂(g) + O₂(g) → 2H₂O(g)

(a) Calculate the volume of hydrogen required to react with 24 dm³ of oxygen.

(b) State the volume of water vapour produced.

Answer

(a)

Volume ratio H₂ : O₂ = 2 : 1

Hydrogen volume = 48 dm³

(b)

2 moles of water vapour are produced, which occupy 48 dm³ at r.t.p.

Question 176 

A solution contains 10 g of sodium chloride dissolved in 0.5 dm³ of solution.

(a) Calculate the concentration in g / dm³.

(b) State what this concentration means.

Answer

(a) Concentration = 10 ÷ 0.5 = 20 g / dm³

(b) Each cubic decimetre of solution contains 20 g of sodium chloride.

Question 177 

25 cm³ of a 1.0 mol / dm³ sodium hydroxide solution reacts with hydrochloric acid.

(a) Convert the volume to dm³.

(b) Calculate the amount of sodium hydroxide present.

Answer

(a) 25 cm³ = 0.025 dm³

(b)

Amount of substance = concentration × volume

= 1.0 × 0.025 = 0.025 mol

Question 178 

Sodium hydroxide reacts with hydrochloric acid:

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

If 0.025 mol of sodium hydroxide reacts:

(a) Calculate the amount of hydrochloric acid required.

(b) Explain how the equation is used.

Answer

(a)

The mole ratio NaOH : HCl is 1 : 1

So 0.025 mol of HCl is required.

(b) The balanced equation gives the mole ratio used to relate reactants.

 

Question 179 

Calcium carbonate reacts with hydrochloric acid to produce carbon dioxide.

(a) Write the balanced symbol equation.

(b) Calculate the volume of carbon dioxide produced from 0.25 mol of calcium carbonate at r.t.p.

Answer

(a)

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

(b)

0.25 mol of CO₂ × 24 = 6 dm³

Question 180 

Describe a step-by-step method for solving a problem involving a solution reacting with a solid to produce a gas.

Answer

First, calculate the amount of substance in solution using concentration and volume.

Second, convert the mass of the solid into moles.

Third, use the balanced equation to identify the limiting reactant.

Fourth, calculate the amount of gas formed from the limiting reactant.

Finally, convert moles of gas into volume using 24 dm³ per mole.

Question 181 

(a) Define the term titration.

(b) State one purpose of carrying out a titration experiment.

Answer

(a) A titration is a laboratory technique used to determine the concentration of a solution by reacting it with another solution of known concentration.

(b) One purpose of a titration is to find the concentration of an unknown solution accurately.

.

Describe the apparatus setup used in a typical acid–alkali titration.

Answer

A measured volume of the solution of unknown concentration is placed in a conical flask using a pipette. An indicator is added to the flask. The solution of known concentration is placed in a burette. The solution from the burette is added slowly to the flask until the indicator changes colour, showing the end point.

Question 183 

Explain why a balanced chemical equation is essential before carrying out any titration calculation.

Answer

A balanced chemical equation shows the mole ratio in which the reactants react. This ratio is essential for converting the moles of one substance into the moles of another. Without the correct equation, titration calculations would be inaccurate.

Question 184 

25.0 cm³ of sodium hydroxide solution is neutralized by 20.0 cm³ of hydrochloric acid of concentration 0.100 mol / dm³.

(a) Convert the volume of acid to dm³.

(b) Calculate the moles of hydrochloric acid used.

Answer

(a) 20.0 cm³ = 0.020 dm³

(b)

Moles of HCl = concentration × volume

= 0.100 × 0.020

= 0.0020 mol

Question 185 

Using the data from Question 184 and the equation:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Calculate the moles of sodium hydroxide present in 25.0 cm³ of solution.

Answer

The equation shows a 1 : 1 mole ratio between HCl and NaOH.

Therefore, moles of NaOH = moles of HCl

= 0.0020 mol

Question 186 

Using the result from Question 185, calculate the concentration of the sodium hydroxide solution in mol / dm³.

Answer

Volume of NaOH = 25.0 cm³ = 0.025 dm³

Concentration = moles ÷ volume

= 0.0020 ÷ 0.025

= 0.080 mol / dm³

Question 187 

Sulfuric acid reacts with sodium hydroxide according to the equation:

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

If 0.010 mol of sodium hydroxide reacts, calculate the moles of sulfuric acid used.

Answer

The mole ratio NaOH : H₂SO₄ is 2 : 1.

Moles of sulfuric acid = 0.010 ÷ 2

= 0.005 mol

Question 188 

A student finds that 0.005 mol of sodium hydroxide is present in 25.0 cm³ of solution.

(a) Convert the volume to dm³.

(b) Calculate the concentration of the solution.

Answer

(a)

25.0 cm³ = 0.025 dm³

(b)

Concentration = 0.005 ÷ 0.025

= 0.20 mol / dm³

Question 189 

A titration requires 0.010 mol of hydrochloric acid. The acid has a concentration of 0.50 mol / dm³.

Calculate the volume of acid required in cm³.

Volume (dm³) = moles ÷ concentration

= 0.010 ÷ 0.50

= 0.020 dm³

0.020 dm³ = 20.0 cm³

Question 190 

Explain the role of an indicator in a titration and name one suitable indicator for an acid–alkali titration.

Answer

An indicator is used to show when the reaction is just complete by changing colour at the end point. This colour change signals that the reactants have reacted in the correct proportions.

One suitable indicator for acid–alkali titrations is phenolphthalein.

Question 191 

(a) Define the term empirical formula.

(b) Define the term molecular formula.

Answer

(a) The empirical formula shows the simplest whole-number ratio of atoms of each element present in a compound.

(b) The molecular formula shows the actual number of atoms of each element present in one molecule of a compound.

Question 192 

Explain why the empirical formula must always be calculated before the molecular formula.

Answer

The empirical formula gives the simplest ratio of atoms in a compound. The molecular formula is always a whole-number multiple of the empirical formula. Without knowing the empirical formula first, it is not possible to determine how many times this ratio occurs in the molecule.

Question 193 

A compound contains 12 g of carbon and 2 g of hydrogen.

Calculate the empirical formula of the compound.

Answer

Carbon: 12 ÷ 12 = 1 mol

Hydrogen: 2 ÷ 1 = 2 mol

Mole ratio = 1 : 2

Empirical formula = CH₂

Question 194 

A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen.

Calculate the empirical formula of the compound.

Answer

Assume 100 g of compound.

Carbon: 40 ÷ 12 = 3.33 mol

Hydrogen: 6.7 ÷ 1 = 6.7 mol

Oxygen: 53.3 ÷ 16 = 3.33 mol

Divide by smallest value (3.33):

C : H : O = 1 : 2 : 1

Empirical formula = CH₂O

Question 195 

Explain how percentage composition data can be treated as mass data when calculating empirical formulae.

Answer

Percentage composition gives the mass of each element in 100 g of the compound. This allows percentages to be treated directly as masses. These masses can then be converted into moles and simplified to find the empirical formula.

Question 196 

A compound has the empirical formula CH₂. Its relative molecular mass is 42.

Calculate the molecular formula of the compound.

Answer

Mass of empirical formula CH₂ = 12 + 2 = 14

Factor = 42 ÷ 14 = 3

Molecular formula = C₃H₆

Question 197 

A compound has an empirical formula mass of 18 and a relative molecular mass of 18.

State and explain the molecular formula.

Answer

Factor = 18 ÷ 18 = 1

This means the molecular formula is the same as the empirical formula.

The molecular formula is H₂O.

Question 198 

During combustion analysis, a compound is found to contain carbon and hydrogen only. Explain how combustion data helps determine the empirical formula.

Answer

During combustion, carbon is converted into carbon dioxide and hydrogen into water. By measuring the masses of carbon dioxide and water produced, the masses of carbon and hydrogen in the original compound can be calculated. These masses are converted into moles and simplified to find the empirical formula.

Question 199 

A compound contains carbon and oxygen in the mole ratio 1 : 1.5.

(a) Explain why this ratio must be changed.

(b) Determine the empirical formula.

Answer

(a) Empirical formulae must contain whole-number ratios because atoms cannot exist in fractions.

(b) Multiply both values by 2:1 : 1.5 → 2 : 3

Empirical formula = C₂O₃

Question 200 

Explain why empirical and molecular formula calculations are important in chemistry.

Answer

These calculations allow chemists to identify unknown compounds, understand chemical composition, relate experimental data to atomic structure, and perform further calculations involving reactions, masses, and moles. They form a key link between laboratory results and chemical theory.

Question 201 

(a) Define percentage yield.

(b) State the formula used to calculate percentage yield.

Answer

(a) Percentage yield is a measure of how much product is obtained from a chemical reaction compared to the maximum amount that could be obtained according to theory.

(b) Percentage yield = (actual yield ÷ theoretical yield) × 100

Question 202 

Explain the difference between theoretical yield and actual yield.

Answer

The theoretical yield is the maximum mass of product that can be calculated from a balanced chemical equation, assuming the reaction goes to completion with no losses. The actual yield is the mass of product that is actually obtained and measured during an experiment. The actual yield is usually lower due to practical losses or side reactions.

Question 203 

A reaction has a theoretical yield of 25 g of product. The actual yield obtained is 20 g.

Calculate the percentage yield.

Answer

Percentage yield = (actual yield ÷ theoretical yield) × 100

= (20 ÷ 25) × 100

= 80%

The percentage yield of the reaction is 80%.

Question 204 

State three reasons why the percentage yield of a reaction is often less than 100%.

Answer

Some reactants may not react completely.

Some product may be lost during filtration, transfer, or purification.

Side reactions may occur, forming unwanted products.

Question 205 

Explain why a percentage yield greater than 100% may sometimes be obtained experimentally.

Answer

A percentage yield greater than 100% may occur if the product is impure. For example, the product may contain unremoved solvent, water, or other impurities. These increase the measured mass of the product, making the actual yield appear greater than the theoretical yield.

Question 206 

Define percentage composition by mass and give the formula used to calculate it.

Answer

Percentage composition by mass shows how much of each element is present in a compound as a percentage of the total mass.

Percentage composition = (mass of element ÷ total mass of compound) × 100

Question 207 

Calculate the percentage composition by mass of hydrogen in water, H₂O.

(Ar: H = 1, O = 16)

Answer

Relative molecular mass of H₂O = (2 × 1) + 16 = 18

Mass of hydrogen = 2

Percentage of hydrogen = (2 ÷ 18) × 100

= 11.1%

The percentage composition of hydrogen in water is 11.1%.

Question 208 

Carbon dioxide has the formula CO₂.

Calculate the percentage composition by mass of oxygen.

(Ar: C = 12, O = 16)

Answer

Relative molecular mass of CO₂ = 12 + (2 × 16) = 44

Mass of oxygen = 32

Percentage of oxygen = (32 ÷ 44) × 100

= 72.7%

The percentage composition of oxygen in carbon dioxide is 72.7%.

Question 209

(a) Define percentage purity.

(b) State the formula used to calculate percentage purity.

Answer

(a) Percentage purity measures how much of a sample is made up of the desired pure substance.

(b) Percentage purity = (mass of pure substance ÷ mass of impure sample) × 100

Question 210 

A 15 g sample of sodium carbonate contains 12 g of pure sodium carbonate.

Calculate the percentage purity of the sample.

Answer

Percentage purity = (mass of pure substance ÷ mass of impure sample) × 100

= (12 ÷ 15) × 100

= 80%

The percentage purity of the sample is 80%.

Thank You!

Sana Shariq

For revision visit

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