CBSE Class 11 Chemistry Some Basic Concepts of Chemistry MCQs Set 05

Answer: (b) 3

Explanation:
1. Zeros at the beginning of a number (0.00) are not significant.
2. The digits 4 and 5 are non-zero, and the zero after them is significant because it comes after the decimal point and to the right of a non-zero digit.
3. So 0.00450 = 4.50 × 10⁻³ has three significant figures.

Teacher's Note:
1. Write the number in scientific notation to see the significant digits clearly.
2. Leading zeros never count, but trailing zeros after a decimal point do.

Answer: (d) 6.4 and 6.2

Explanation:
1. When the digit to be dropped is exactly 5, the preceding digit is raised by one only if it is odd, and left unchanged if it is even.
2. In 6.35 the preceding digit 3 is odd, so it becomes 4 and the result is 6.4.
3. In 6.25 the preceding digit 2 is even, so it stays and the result is 6.2.

Teacher's Note:
1. Odd digit goes up, even digit stays when the dropped digit is 5.
2. Check each number separately.

Answer: (a) 26.85 °C

Explanation:
1. The two scales are related by K = °C + 273.15.
2. So °C = K − 273.15 = 300 − 273.15 = 26.85 °C.

Teacher's Note:
1. Kelvin is the SI unit, so a Kelvin value is always larger than the Celsius value for ordinary temperatures.
2. Subtract, do not add, when going from K to °C.

Answer: (c) Sand with common salt and chalk powder in water

Explanation:
1. A heterogeneous mixture has visible boundaries between its parts (phases).
2. Sand with salt and chalk powder in water are both like that.
3. Air, brass, sea water, gasoline and steel have the same composition throughout, so they are homogeneous mixtures.

Teacher's Note:
1. Ask whether you can see different parts with the naked eye.
2. Solutions and alloys are homogeneous.

Answer: (d) multiple proportions

Explanation:
1. The mass of carbon is fixed at 12 g in both compounds.
2. The masses of oxygen, 16 g and 32 g, are in the simple ratio 1 : 2.
3. Two elements forming more than one compound with masses in a simple whole-number ratio is the law of multiple proportions.

Teacher's Note:
1. Look for one fixed mass of an element and a simple ratio of the other.
2. Dalton proposed this law in 1803.

Answer: (b) Gay-Lussac's law of gaseous volumes

Explanation:
1. The volumes of the reacting gases and the product are in the simple ratio 10 : 10 : 20, that is 1 : 1 : 2.
2. Gases combine in simple whole-number volume ratios at the same temperature and pressure, which is Gay-Lussac's law.

Teacher's Note:
1. Volumes of gases, not masses, are being compared.
2. Avogadro's law later explained this law in terms of molecules.

Answer: (c) 1.9927 × 10⁻²³ g

Explanation:
1. One mole of carbon-12 atoms has a mass of 12 g and contains 6.022 × 10²³ atoms.
2. Mass of one atom = 12 g ÷ 6.022 × 10²³ = 1.9927 × 10⁻²³ g.

Teacher's Note:
1. Divide the molar mass by Avogadro's number to get the mass of one particle.
2. 1.66 × 10⁻²⁴ g is one-twelfth of this, i.e. 1 u.

Answer: (a) 7.3 × 10⁻²³ g

Explanation:
1. The molecular mass of CO₂ = 12 + 2 × 16 = 44 u.
2. Mass of one molecule = 44 × 1.66 × 10⁻²⁴ g = 7.3 × 10⁻²³ g.

Teacher's Note:
1. First find the molecular mass in u, then convert with 1 u = 1.66 × 10⁻²⁴ g.
2. Dividing 44 g by 6.022 × 10²³ gives the same answer.

Answer: (a) 0.1

Explanation:
1. Molar mass of SO₂Cl₂ = 32 + 2 × 16 + 2 × 35.5 = 135 g mol⁻¹.
2. Moles = 13.5 ÷ 135 = 0.1.

Teacher's Note:
1. Add the atomic masses of every atom in the formula.
2. Moles = mass ÷ molar mass.

Answer: (d) 1.5 × 10²³

Explanation:
1. Molar mass of NH₃ = 14 + 3 = 17 g mol⁻¹.
2. Moles = 4.25 ÷ 17 = 0.25 mol.
3. Molecules = 0.25 × 6.022 × 10²³ = 1.5 × 10²³.

Teacher's Note:
1. Convert mass to moles first, then multiply by Avogadro's number.
2. 0.25 mol is one-quarter of 6.022 × 10²³.

Answer: (b) 22.4 L of CO₂ at STP

Explanation:
1. 22.4 L of CO₂ at STP is 1 mol, which has 3 mol of atoms.
2. 6.022 × 10²¹ molecules is 0.01 mol, giving 0.03 mol of atoms; 0.44 g of CO₂ is also 0.01 mol, giving 0.03 mol of atoms.
3. 0.1 mol of CO has 0.2 mol of atoms, so 22.4 L of CO₂ has the most.

Teacher's Note:
1. Convert every quantity to moles of atoms before comparing.
2. Remember that CO₂ has three atoms per molecule.

Answer: (c) 9.6 kg

Explanation:
1. C₂H₄ + 3O₂ → 2CO₂ + 2H₂O; 28 g of C₂H₄ needs 96 g of O₂.
2. 2.8 kg is 2800 g = 100 mol of C₂H₄, which needs 300 mol of O₂.
3. Mass of O₂ = 300 × 32 = 9600 g = 9.6 kg.

Teacher's Note:
1. Write and balance the equation before using the mole ratio.
2. kg and g work equally well if you stay consistent.

Answer: (d) 7.96 g

Explanation:
1. Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃.
2. Moles of Pb(NO₃)₂ = 10.2 ÷ 331 = 0.0308 mol, and of KI = 5.73 ÷ 166 = 0.0345 mol.
3. 0.0308 mol of Pb(NO₃)₂ needs 0.0616 mol of KI, so KI is the limiting reagent.
4. PbI₂ formed = 0.0345 ÷ 2 = 0.01726 mol, and the mass is 0.01726 × 461 = 7.96 g.

Teacher's Note:
1. Always compare moles with the coefficients to find the limiting reagent.
2. The product depends on the limiting reagent only.

Answer: (c) 4

Explanation:
1. O₂ used = 9 − 3 = 6 mol.
2. From the equation, 3 mol of O₂ gives 2 mol of N₂O₃, so 6 mol of O₂ gives 4 mol of N₂O₃.

Teacher's Note:
1. Find how much of the reactant is used up, not how much is left.
2. Use the coefficient ratio 3 : 2.

Answer: (a) MO

Explanation:
1. Per 100 g: M = 60 g and O = 40 g.
2. Moles of M = 60 ÷ 24 = 2.5 and moles of O = 40 ÷ 16 = 2.5.
3. The ratio is 1 : 1, so the empirical formula is MO.

Teacher's Note:
1. Divide percentage by atomic mass, then divide by the smallest value.
2. A 1 : 1 ratio needs no further multiplying.

Answer: (b) 1.00 M

Explanation:
1. Moles of HCl = 0.050 × 2.00 + 0.100 × 1.00 + 0.100 × 0.500 = 0.100 + 0.100 + 0.050 = 0.250 mol.
2. Total volume = 250 mL = 0.250 L.
3. Molarity = 0.250 ÷ 0.250 = 1.00 M.

Teacher's Note:
1. Add the moles, not the molarities.
2. Add the volumes to get the final volume.

Answer: (c) 3.2 m

Explanation:
1. Take 1 mol of solution: I₂ = 0.2 mol and C₆H₆ = 0.8 mol.
2. Mass of benzene = 0.8 × 78 = 62.4 g = 0.0624 kg.
3. Molality = 0.2 ÷ 0.0624 = 3.2 mol kg⁻¹.

Teacher's Note:
1. Molality needs the mass of solvent in kg.
2. Choose a convenient total of 1 mol when mole fraction is given.

Answer: (a) 50 g of MgO and 10 g of O₂

Explanation:
1. 2Mg + O₂ → 2MgO. 30 g of Mg is 1.25 mol and needs 0.625 mol of O₂ (20 g).
2. So Mg is completely used up and 30 − 20 = 10 g of O₂ is left over.
3. MgO formed = 1.25 mol × 40 g/mol = 50 g (this equals 30 g + 20 g, as mass is conserved).

Teacher's Note:
1. Find the limiting reagent first.
2. Check the answer: 50 g + 10 g = 60 g, the total mass taken.

Answer: (d) Molar mass is the mass of one molecule

Explanation:
1. Molar mass is the mass of 6.022 × 10²³ units of a substance, expressed in g mol⁻¹.
2. It is the molecular (or atomic) mass expressed in grams, not the mass of a single molecule.
3. The mass of one molecule is the molar mass divided by Avogadro's number.

Teacher's Note:
1. Do not confuse molar mass with the mass of a single particle.
2. Use the unit g mol⁻¹ for molar mass.

Answer: (a) Molality

Explanation:
1. Molality is moles of solute per kilogram of solvent, and mass does not change with temperature.
2. Molarity and other terms that use the volume of solution change because the volume expands or contracts on heating or cooling.

Teacher's Note:
1. Terms with volume in the denominator depend on temperature.
2. Mass percent, mole fraction and molality are temperature independent.