CBSE Class 9 Chemistry Atoms and Molecules MCQs Set 06

Answer: (b) 22.4 g

Explanation:
1. By the law of conservation of mass, total mass of reactants = total mass of products.
2. 20 + 10.6 = 30.6 g, and 30.6 - 8.2 = 22.4 g.
3. So the mass of \( \ce{BaSO4} \) is 22.4 g.

Teacher's Note:
1. Mass is neither created nor destroyed in a chemical reaction.
2. Subtract the known product from the total reactant mass.

Answer: (b) 11.11 percent

Explanation:
1. Molar mass of \( \ce{H2O} \) = 2 + 16 = 18 g.
2. Percentage of hydrogen = 2/18 x 100 = 11.11 percent.

Teacher's Note:
1. Percentage by mass is mass of element divided by mass of compound times 100.
2. Oxygen is 88.89 percent.

Answer: (c) 47.2

Explanation:
1. Molar mass of \( \ce{CaSO4} \) = 40 + 32 + 64 = 136.
2. Percentage of oxygen = 64/136 x 100 = 47.06, which is closest to 47.2.

Teacher's Note:
1. Multiply the atomic mass of oxygen by 4 for four oxygen atoms.
2. Divide by the molar mass of the compound.

Answer: (c) Statement I is true but Statement II is false

Explanation:
1. 5.6 + 4.4 = 10 g, so the data are consistent and Statement I is true.
2. The data show conservation of mass, not constant proportions, so Statement II is false.

Teacher's Note:
1. Constant proportions needs two samples of the same compound with the same mass ratio.
2. Here the total mass is conserved.

Answer: (d) 12 : 16 and 12 : 32

Explanation:
1. In CO, 12 g of carbon combines with 16 g of oxygen.
2. In \( \ce{CO2} \), 12 g of carbon combines with 32 g of oxygen.
3. So the ratios are 12 : 16 and 12 : 32.

Teacher's Note:
1. Use atomic masses C = 12 and O = 16.
2. The oxygen masses 16 and 32 are in the simple ratio 1 : 2.

Answer: (d) Law of multiple proportions

Explanation:
1. The masses of M that combine with a fixed mass of Cl are 0.89 and 1.791.
2. Their ratio is about 1 : 2, a simple whole number ratio, which is the law of multiple proportions.

Teacher's Note:
1. Two elements forming more than one compound follow this law.
2. Fix the mass of one element and compare the other.

Answer: (d) 16.0 g

Explanation:
1. In \( \ce{CS2} \) the ratio of masses is C : S = 12 : 64.
2. For 3.0 g of carbon, sulphur = 3.0 x 64/12 = 16.0 g.

Teacher's Note:
1. Find the mass ratio from the formula first.
2. Then scale to the given mass.

Answer: (b) Law of definite proportions

Explanation:
1. In the first sample C : O = 1.2 : 3.2 = 0.375.
2. In the second sample 27.3 : 72.7 = 0.375.
3. The same compound has the same ratio, so the law of definite proportions holds.

Teacher's Note:
1. Same compound, same composition, whatever the source.
2. Compare ratios of masses.

Answer: (b) Berzelius

Explanation:
1. Berzelius suggested using one or two letters of the name of the element as its symbol.
2. Dalton had used pictorial symbols.

Teacher's Note:
1. Symbols come from English or Latin names.
2. For example Na from natrium.

Answer: (c) Sodium

Explanation:
1. Natrium is the Latin name of sodium, so its symbol is Na.

Teacher's Note:
1. Kalium gives K for potassium.
2. Ferrum gives Fe for iron.

Answer: (b) 11

Explanation:
1. Atomicity is the number of atoms in a molecule.
2. 2 K + 2 Cr + 7 O = 11 atoms.

Teacher's Note:
1. Add the subscripts of all elements.
2. Atomicity of \( \ce{O2} \) is 2 and of \( \ce{P4} \) is 4.

Answer: (a) Atom

Explanation:
1. An atom is not destroyed in chemical reactions, only rearranged.
2. Molecules and compounds change in chemical reactions.

Teacher's Note:
1. Atoms are conserved in all chemical changes.
2. Molecules are not.

Answer: (c) One mole of hydrogen gas contains Avogadro's number of atoms

Explanation:
1. One mole of hydrogen gas, \( \ce{H2} \), contains Avogadro's number of molecules.
2. Since each molecule has 2 atoms, it contains twice that number of atoms.

Teacher's Note:
1. Mole refers to the particles named, here molecules of \( \ce{H2} \).
2. Do not confuse atoms with molecules.

Answer: (c) a charged particle

Explanation:
1. An ion is an atom or group of atoms with a positive or negative electric charge.

Teacher's Note:
1. Cation is positive and anion is negative.
2. Ions form by loss or gain of electrons.

Answer: (c) Sulphate

Explanation:
1. Sulphate, \( \ce{SO4^2-} \), is a group of atoms carrying a charge.
2. Sulphide, chloride and nitride are single atom (monatomic) ions.

Teacher's Note:
1. Polyatomic ions have more than one atom.
2. Others are ammonium, nitrate and carbonate.

Answer: (c) there are fewer electrons than protons

Explanation:
1. When an atom loses electrons, protons outnumber electrons, so the particle has a net positive charge.
2. Neutrons carry no charge.

Teacher's Note:
1. Positive ion means loss of electrons.
2. Number of protons does not change.

Answer: (c) \( \ce{MCl2} \) and \( \ce{MSO4} \)

Explanation:
1. Hydrogen phosphate, \( \ce{HPO4^2-} \), has valency 2, so M has valency 2.
2. The chloride is \( \ce{MCl2} \) and the sulphate is \( \ce{MSO4} \).

Teacher's Note:
1. Find the metal's valency from the given compound.
2. Then use the criss-cross method.

Answer: (d) \( \ce{Ca3(PO4)2} \)

Explanation:
1. Calcium is \( \ce{Ca^2+} \) and phosphate is \( \ce{PO4^3-} \).
2. Criss-cross: 3 calcium ions and 2 phosphate ions, so \( \ce{Ca3(PO4)2} \).

Teacher's Note:
1. Put a bracket round a polyatomic ion when it needs a subscript.
2. Check that the charges balance: 6 plus and 6 minus.

Answer: (a) a molecule contains 2 atoms of P and 5 atoms of O

Explanation:
1. The subscripts give the number of atoms in a molecule: 2 atoms of phosphorus and 5 atoms of oxygen.
2. The mass ratio would be 62 : 80.

Teacher's Note:
1. Subscripts show atoms, not mass.
2. Mass ratio needs atomic masses.

Answer: (a) \( \ce{MCrO4} \)

Explanation:
1. Chromate is \( \ce{CrO4^2-} \) with valency 2.
2. A divalent metal M has valency 2, so the formula is \( \ce{MCrO4} \).

Teacher's Note:
1. Equal valencies cancel out.
2. Chromate carries two negative charges.

Answer: (d) Statement I is false but Statement II is true

Explanation:
1. Oxygen exists as \( \ce{O2} \), so its molecular mass is 32 u, which makes Statement I false.
2. Its atomic mass is 16 u, so Statement II is true.

Teacher's Note:
1. Molecular mass = sum of atomic masses in the molecule.
2. Oxygen gas is diatomic.

Answer: (a) mole

Explanation:
1. The mole is the SI unit for the amount of substance.
2. Kilogram measures mass and second measures time.

Teacher's Note:
1. One mole contains the Avogadro number of particles.
2. Avogadro number is \( \ce{6.022 x 10^23} \).

Answer: (d) 24 g

Explanation:
1. Molar mass of \( \ce{CaCO3} \) = 100 g per mol, so 50 g = 0.5 mol.
2. Each mole has 3 mol of O atoms = 48 g, so 0.5 mol has 24 g of oxygen.

Teacher's Note:
1. Oxygen is 48 percent of \( \ce{CaCO3} \) by mass.
2. 48 percent of 50 g is 24 g.

Answer: (a) \( \ce{6.68 x 10^23} \)

Explanation:
1. Sucrose \( \ce{C12H22O11} \) has molar mass 342 g, so 3.42 g = 0.01 mol, giving 0.11 mol of O atoms = \( \ce{0.662 x 10^23} \) atoms.
2. 18 g water is 1 mol, giving \( \ce{6.022 x 10^23} \) O atoms.
3. Total = \( \ce{6.68 x 10^23} \).

Teacher's Note:
1. Do not forget the oxygen in the water.
2. Add the contributions from both substances.

Answer: (b) x/2 = y = z = w/4

Explanation:
1. I: 0.5 mol of \( \ce{N2} \) = 0.5 x 28 = 14 g.
2. II: 0.5 mol of N atoms = 0.5 x 14 = 7 g.
3. III: \( \ce{3.011 x 10^23} \) atoms is 0.5 mol of atoms = 7 g.
4. IV: 1 mol of \( \ce{N2} \) = 28 g.
5. So x = 14, y = 7, z = 7, w = 28, giving x/2 = y = z = w/4.

Teacher's Note:
1. Distinguish moles of molecules from moles of atoms.
2. Nitrogen gas has 28 g per mole of molecules.

Answer: (b) W = 55.8, X = 32, Y = 1.008, Z = 58.5

Explanation:
1. One mole of iron atoms has mass 55.8 g and one mole of \( \ce{O2} \) molecules has mass 32 g.
2. One mole of hydrogen atoms has mass 1.008 g and one mole of NaCl has mass 23 + 35.5 = 58.5 g.

Teacher's Note:
1. Molar mass in grams equals atomic or molecular mass in u.
2. Oxygen gas is \( \ce{O2} \), not O.

Answer: (d) 0.4 gram atoms, 0.1 gram molecules

Explanation:
1. Atomic mass of P = 31, so gram atoms = 12.4/31 = 0.4.
2. Phosphorus exists as \( \ce{P4} \) (molar mass 124), so gram molecules = 12.4/124 = 0.1.

Teacher's Note:
1. Phosphorus molecule has four atoms.
2. Gram atoms are four times gram molecules.

Answer: (d) 2 g of \( \ce{H2} \)

Explanation:
1. Moles: 8/16 = 0.5, 4.4/44 = 0.1, 34.2/342 = 0.1 and 2/2 = 1.
2. The largest number of moles, 1 mol of \( \ce{H2} \), means the largest number of particles.

Teacher's Note:
1. Same particle type here, so more moles means more particles.
2. Convert each mass to moles.

Answer: (a) 1 : 1

Explanation:
1. 25.4 g of \( \ce{I2} \) = 0.1 mol, giving 0.2 mol of I atoms.
2. 14.2 g of \( \ce{Cl2} \) = 0.2 mol, giving 0.4 mol of Cl atoms.
3. Let x mol ICl and y mol \( \ce{ICl3} \): x + y = 0.2 and x + 3y = 0.4, so y = 0.1 and x = 0.1.
4. The ratio is 1 : 1.

Teacher's Note:
1. Write mole balance of each element.
2. Solve the two simple equations.