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The Vault
RowQ
The Vault
CBSE Class 9 Chemistry · 12 questions · 29 marks
Atoms are far too small to be seen or weighed one at a time, yet chemists routinely measure them out by the trillion with everyday laboratory balances. This chapter shows you how they manage it — through the laws of chemical combination, Dalton's atomic theory, relative atomic masses, and above all the mole, the counting unit that links the mass you weigh on a balance to the number of particles you actually have.
A sample of pure copper oxide prepared in a Delhi laboratory and another prepared in a Chennai laboratory by a completely different method are both found to contain copper and oxygen in the mass ratio 4 : 1. This observation is a direct illustration of:
Answer
The law of constant proportions. This law states that a given pure compound always contains the same elements combined in the same fixed ratio by mass, regardless of its source or method of preparation — which is exactly what the identical 4 : 1 ratio in two independently prepared samples demonstrates. The law of conservation of mass concerns the total mass before and after a reaction, not the composition of a compound, so it does not apply here.
The molar mass of calcium carbonate, CaCO₃, is (atomic masses: Ca = 40 u, C = 12 u, O = 16 u):
Answer
100 g/mol. Adding the atomic masses: Ca = 40, C = 12, and 3 oxygen atoms = 3 × 16 = 48. The formula unit mass is 40 + 12 + 48 = 100 u, so the molar mass is 100 g/mol. The commonest mistake is to use only one oxygen atom instead of three, which wrongly gives 68.
The number of molecules present in 0.5 mole of carbon dioxide gas is:
Answer
3.011 × 10²³. One mole of any substance contains 6.022 × 10²³ particles, so 0.5 mole contains 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules of CO₂. Note that the number of atoms would be three times this, since each CO₂ molecule contains three atoms.
The correct chemical formula of aluminium sulphate is:
Answer
Al₂(SO₄)₃. Aluminium has a valency of 3 and the sulphate ion, SO₄, has a valency of 2. Criss-crossing the valencies puts 2 as the subscript of aluminium and 3 as the subscript of sulphate. Because more than one sulphate ion is needed, the whole polyatomic ion must be enclosed in brackets before the subscript is applied, giving Al₂(SO₄)₃.
Assertion (A): A 22 g sample of carbon dioxide contains 0.5 mole of CO₂ molecules. Reason (R): The molar mass of carbon dioxide is 44 g/mol, and the number of moles equals the given mass divided by the molar mass.
Answer
Both A and R are true and R is the correct explanation of A. The molar mass of CO₂ is 12 + (2 × 16) = 44 g/mol, so n = given mass ÷ molar mass = 22 ÷ 44 = 0.5 mole. The relationship stated in R is precisely the calculation that produces the value claimed in A, so R explains A completely.
State the law of conservation of mass. In a sealed flask, 8.4 g of a metal carbonate is heated and decomposes into 4.4 g of a metal oxide and a gas. Calculate the mass of gas produced and explain which law your calculation relies on.
Answer
The law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction, so in a closed system the total mass of the products equals the total mass of the reactants. Mass of gas = mass of reactant − mass of solid product = 8.4 g − 4.4 g = 4.0 g of gas. The calculation relies directly on the law of conservation of mass: because the flask was sealed, no matter could escape, so the missing 4.0 g must be accounted for by the gas that was formed.
Calculate the number of moles and the number of formula units present in 60 g of sodium hydroxide, NaOH. (Atomic masses: Na = 23 u, O = 16 u, H = 1 u.)
Answer
Step 1 — Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Step 2 — Number of moles, n = given mass ÷ molar mass = 60 ÷ 40 = 1.5 mol. Step 3 — Number of formula units = n × 6.022 × 10²³ = 1.5 × 6.022 × 10²³ = 9.033 × 10²³ formula units of NaOH. So 60 g of sodium hydroxide is 1.5 moles and contains 9.033 × 10²³ formula units.
Define one atomic mass unit (1 u). Using it, calculate the molecular mass of nitric acid, HNO₃. (Atomic masses: H = 1 u, N = 14 u, O = 16 u.)
Answer
One atomic mass unit is defined as exactly one-twelfth of the mass of one atom of carbon-12. All atomic and molecular masses are relative values measured against this agreed standard. Molecular mass of HNO₃ = mass of H + mass of N + (3 × mass of O) = 1 + 14 + 48 = 63 u. Its molar mass is therefore 63 g/mol.
Calculate the mass in grams of 3.011 × 10²² atoms of magnesium. (Atomic mass of Mg = 24 u.)
Answer
Step 1 — Convert the number of atoms into moles: n = number of atoms ÷ 6.022 × 10²³ = (3.011 × 10²²) ÷ (6.022 × 10²³) = 0.05 mol. Step 2 — Molar mass of magnesium = 24 g/mol. Step 3 — Mass = n × molar mass = 0.05 × 24 = 1.2 g. So 3.011 × 10²² atoms of magnesium weigh 1.2 g — a reminder that even a mass small enough to sit on a fingertip contains tens of thousands of billions of billions of atoms.
State the main postulates of Dalton's atomic theory. Explain how the theory accounts for the law of conservation of mass and the law of constant proportions, and state two of its limitations.
Answer
Postulates: (1) All matter is made up of extremely small particles called atoms. (2) Atoms are indivisible and cannot be created or destroyed in a chemical reaction. (3) All atoms of a given element are identical in mass and in chemical properties. (4) Atoms of different elements have different masses and different chemical properties. (5) Atoms of different elements combine in small whole-number ratios to form compounds. (6) In a given compound, the relative number and kinds of atoms are constant. Explanation of the law of conservation of mass: Since atoms are neither created nor destroyed during a chemical reaction, and each atom carries a fixed mass, the same atoms simply rearrange themselves into new combinations. The total number of atoms of each element before and after the reaction is identical, so the total mass cannot change — which is exactly what the law asserts. Explanation of the law of constant proportions: Since a particular compound always contains the same kinds of atoms combined in the same fixed whole-number ratio, and each kind of atom has its own fixed mass, the ratio by mass of the elements in that compound must also be fixed. Water always contains two hydrogen atoms for every oxygen atom, so it always shows a 1 : 8 mass ratio of hydrogen to oxygen, whatever its source. Limitations: (1) The postulate that atoms are indivisible was later disproved, since atoms are made up of still smaller particles — electrons, protons and neutrons. (2) The postulate that all atoms of an element are identical in mass fails for elements with isotopes, whose atoms have different masses; conversely, isobars show that atoms of different elements can have the same mass. Dalton's theory also gave no explanation of why atoms combine at all, or of the forces holding them together.
A fertiliser bag lists ammonium sulphate, (NH₄)₂SO₄, as its active ingredient. For a 33 g sample of pure ammonium sulphate, calculate (i) its molar mass, (ii) the number of moles present, (iii) the number of formula units present, and (iv) the number of nitrogen atoms present. (Atomic masses: N = 14 u, H = 1 u, S = 32 u, O = 16 u.)
Answer
(i) Molar mass of (NH₄)₂SO₄: the bracket contains NH₄ and is taken twice, so there are 2 nitrogen atoms and 8 hydrogen atoms, plus 1 sulphur atom and 4 oxygen atoms. N: 2 × 14 = 28 H: 8 × 1 = 8 S: 1 × 32 = 32 O: 4 × 16 = 64 Total = 28 + 8 + 32 + 64 = 132 g/mol. (ii) Number of moles, n = given mass ÷ molar mass = 33 ÷ 132 = 0.25 mol. (iii) Number of formula units = n × 6.022 × 10²³ = 0.25 × 6.022 × 10²³ = 1.5055 × 10²³ formula units. (iv) Each formula unit contains 2 nitrogen atoms, so number of N atoms = 2 × 1.5055 × 10²³ = 3.011 × 10²³ atoms of nitrogen. As a check on part (iv), the same answer can be reached through moles: 0.25 mol of the compound contains 0.5 mol of nitrogen atoms, and 0.5 × 6.022 × 10²³ = 3.011 × 10²³, which agrees. This kind of calculation is what allows a manufacturer to state the exact nitrogen content of a fertiliser from the mass of the compound alone.
Ravi carries out a reaction in a conical flask fitted with a tight stopper. He mixes a solution containing 5.2 g of barium chloride with a solution containing 3.55 g of sodium sulphate. A dense white precipitate of barium sulphate forms at once, leaving sodium chloride dissolved in the liquid. He weighs the sealed flask and its contents before and after the reaction. Answer the following: (a) Which law is Ravi testing, and what should the two weighings show? (b) Write the total mass of the products he should expect. (c) His friend repeats the experiment in an open beaker with a reaction that releases a gas, and finds the mass has decreased. Has the law been broken? Explain. (d) Why must the flask be stoppered for this experiment to be a fair test?
Answer
(a) Ravi is testing the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. The two weighings should be identical: the mass of the sealed flask and its contents after the reaction should equal the mass before it, showing that forming a precipitate has not changed the total quantity of matter present. (b) Total mass of reactants = 5.2 g + 3.55 g = 8.75 g. Since no matter can enter or leave the sealed flask, the total mass of the products — the barium sulphate precipitate plus the dissolved sodium chloride — must also be 8.75 g. (c) No, the law has not been broken. In an open beaker the gas produced escapes into the atmosphere and is therefore no longer on the balance, so the recorded mass falls. If the escaping gas were collected and weighed, the total mass of all the products would still equal the total mass of the reactants. The law applies to a closed system. (d) The flask must be stoppered so that nothing can enter or leave during the reaction. A stopper keeps the system closed, which guarantees that any change in the reading of the balance is caused by the chemical reaction itself and not by matter escaping as vapour or gas, or by air and moisture getting in. Only then is the comparison of the before and after masses a fair test of the law.
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