JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 12

Answer: (b) \( 1.81\times10^{24} \) molecules and \( 5.42\times10^{24} \) atoms

Explanation:
1. Moles of \( \ce{H2O} \) = 54/18 = 3 mol.
2. Molecules = 3 x \( 6.022\times10^{23} = 1.81\times10^{24} \).
3. Each molecule has 3 atoms, so atoms = \( 3\times1.81\times10^{24} = 5.42\times10^{24} \).

Teacher's Note:
1. Atoms are always molecules multiplied by atomicity. Do not mix up the two counts.

Answer: (b) \( 3.61\times10^{24} \)

Explanation:
1. Moles of \( \ce{CO2} \) = 88/44 = 2 mol.
2. Molecules = \( 2\times6.022\times10^{23} = 1.2\times10^{24} \).
3. Each molecule has 3 atoms, so atoms = \( 3.61\times10^{24} \).

Teacher's Note:
1. Count atoms per molecule (1 C + 2 O = 3) and multiply by the number of molecules.

Answer: (d) \( 1.99\times10^{-23} \)

Explanation:
1. One mole of \( ^{12}\text{C} \) weighs 12 g and holds \( N_A \) atoms.
2. Mass of one atom = \( 12/6.022\times10^{23} = 1.99\times10^{-23} \) g.

Teacher's Note:
1. Mass of one atom = molar mass divided by Avogadro number. This is 12 times the value of 1 amu in grams.

Answer: (b) 3.2 g

Explanation:
1. Molar mass of \( \ce{CH3COOH} \) = 60 g/mol, so 6 g = 0.1 mol.
2. Each molecule has 2 oxygen atoms, so O atoms = 0.2 mol.
3. Mass of O = 0.2 x 16 = 3.2 g.

Teacher's Note:
1. Find moles of the compound, multiply by atoms per formula unit, and then by the atomic mass.

Answer: (a) 180 g

Explanation:
1. Molar mass of \( \ce{CuSO4.5H2O} \) = 63.5 + 32 + 64 + 90 = 249.5 g/mol.
2. Moles = 499/249.5 = 2 mol.
3. Water = 2 x 5 = 10 mol = 180 g.

Teacher's Note:
1. The 5 in the formula gives moles of water per mole of the salt. Convert moles to grams at the end.

Answer: (d) 2.44 g

Explanation:
1. Oxygen atoms per formula unit = 4 + 7 = 11.
2. Moles of O atoms = \( 6.022\times10^{22}/6.022\times10^{23} = 0.1 \) mol.
3. Moles of salt = 0.1/11. Molar mass = 142 + 126 = 268 g/mol.
4. Mass = 268 x 0.1/11 = 2.44 g.

Teacher's Note:
1. Count oxygen from both the sulphate and the water of crystallisation. Then divide the atom count by 11.

Answer: (c) 948 g

Explanation:
1. Oxygen atoms per formula unit = 4 + 12 + 24 = 40.
2. 0.64 kg = 640 g of O = 40 mol of O atoms.
3. So 1 mol of alum is needed. Molar mass = 78 + 32 + 64 + 54 + 288 + 432 = 948 g/mol.
4. Mass = 948 g.

Teacher's Note:
1. Add up oxygen from all three parts of the double salt. Here the moles of O equal the atoms per formula unit, so exactly one mole is needed.

Answer: (a) 33.6 L

Explanation:
1. Moles of \( \ce{H3PO4} \) = 98/98 = 1 mol, which has 3 mol of H atoms.
2. Moles of \( \ce{H2} \) = 3/2 = 1.5 mol.
3. Volume at STP = 1.5 x 22.4 = 33.6 L.

Teacher's Note:
1. Equate the moles of hydrogen atoms, not moles of molecules. One mole of \( \ce{H2} \) has two moles of H atoms.

Answer: (a) 2

Explanation:
1. n = PV/RT = (2 x 14)/(0.0821 x 273) = 1.25 mol.
2. Molar mass = 80/1.25 = 64 g/mol.
3. 32 + 16x = 64, so x = 2.

Teacher's Note:
1. Use the ideal gas equation to get moles, then molar mass = mass/moles, and fit the formula.

Answer: (d) 100

Explanation:
1. n = PV/RT = (1 x 0.0048)/(0.08 x 300) = 2 x \( 10^{-4} \) mol.
2. Molar mass of \( X_2 \) = 0.040/(2 x \( 10^{-4} \)) = 200 g/mol.
3. Atomic mass of X = 200/2 = 100.

Teacher's Note:
1. The substance is diatomic, so halve the molar mass to get the atomic mass. Convert mL to L and degrees C to K first.

Answer: (b) 27.6 g

Explanation:
1. Moles of \( \ce{HNO3} \) = 25.2/63 = 0.4 mol.
2. \( \ce{NO2} \) needed = 0.4 x 3/2 = 0.6 mol.
3. Mass = 0.6 x 46 = 27.6 g.

Teacher's Note:
1. Use the coefficient ratio 3 : 2 between \( \ce{NO2} \) and \( \ce{HNO3} \).

Answer: (a) \( 3.6\times10^{19} \)

Explanation:
1. Moles of \( \ce{UF6} \) = \( 7.04\times10^{-3}/352 = 2\times10^{-5} \) mol.
2. Moles of \( \ce{F2} \) = 3 x \( 2\times10^{-5} = 6\times10^{-5} \) mol.
3. Molecules = \( 6\times10^{-5}\times6\times10^{23} = 3.6\times10^{19} \).

Teacher's Note:
1. Convert mg to g, use the 1 : 3 ratio, and finally multiply by Avogadro number.

Answer: (a) 63 L

Explanation:
1. Moles of \( \ce{NH4NO3} \) = 16/80 = 0.2 mol.
2. Gas moles = 0.2 x (2 + 1 + 4)/2 = 0.7 mol (water is also a gas here).
3. T = 900 K. V = nRT/P = (0.7 x 0.0821 x 900)/0.821 = 63 L.

Teacher's Note:
1. Count every gaseous product, including water vapour at this temperature. Use the absolute temperature in K.

Answer: (a) 58.8 g

Explanation:
1. Molar mass of \( \ce{H4P2O7} \) = 178 g/mol, so 53.4 g = 0.3 mol.
2. \( \ce{H3PO4} \) needed = 2 x 0.3 = 0.6 mol.
3. Mass = 0.6 x 98 = 58.8 g.

Teacher's Note:
1. Two moles of \( \ce{H3PO4} \) give one mole of the product, so double the moles.

Answer: (a) 154 g of \( \ce{CCl4} \); 24 g of C left

Explanation:
1. \( \ce{C + 2Cl2 -> CCl4} \). Moles of C = 3, moles of \( \ce{Cl2} \) = 142/71 = 2.
2. Dividing by coefficients: C = 3, \( \ce{Cl2} \) = 1. So \( \ce{Cl2} \) is the limiting reagent.
3. \( \ce{CCl4} \) formed = 2/2 = 1 mol = 154 g.
4. C used = 1 mol, so 2 mol = 24 g of carbon is left.

Teacher's Note:
1. Divide each mole count by its coefficient. The smaller value marks the limiting reagent.

Answer: (b) 0.1185 mol

Explanation:
1. Balanced: \( \ce{4KO2 + 2H2O -> 4KOH + 3O2} \).
2. \( \ce{KO2} \)/4 = 0.0395 and \( \ce{H2O} \)/2 = 0.05, so \( \ce{KO2} \) is limiting.
3. \( \ce{O2} \) = 0.158 x 3/4 = 0.1185 mol.

Teacher's Note:
1. The equation in the book is unbalanced. After balancing, compare moles divided by coefficients.

Answer: (b) 0.25 mol

Explanation:
1. LiH/6 = 0.333 and \( \ce{BF3} \)/8 = 0.25, so \( \ce{BF3} \) is limiting.
2. \( \ce{B2H6} \) = 2.0/8 = 0.25 mol.

Teacher's Note:
1. The reagent with the smaller (moles / coefficient) value limits the product.

Answer: (d) 9.5 g

Explanation:
1. Moles: \( \ce{TiO2} \) = 0.054, C = 0.48, \( \ce{Cl2} \) = 0.10.
2. Moles / coefficient: 0.018, 0.12, 0.0167, so \( \ce{Cl2} \) is limiting.
3. \( \ce{TiCl4} \) = 0.10 x 3/6 = 0.05 mol x 190 = 9.5 g.

Teacher's Note:
1. Check all three reactants. The limiting one is not always the one with the least mass or least moles.

Answer: (c) 40% Mg and 60% Al

Explanation:
1. \( \ce{H2} \) = 1.12/22.4 = 0.05 mol. Mg gives 1 mol \( \ce{H2} \) per mol; Al gives 1.5 mol per mol.
2. Let Mg = x g: x/24 + 1.5(1 - x)/27 = 0.05.
3. Solving gives x = 0.4 g, so Mg is 40% and Al is 60%.

Teacher's Note:
1. Write one equation for total hydrogen using each metal's hydrogen yield per mole.

Answer: (b) 75%

Explanation:
1. Moles of CaO = 0.84/56 = 0.015 mol = moles of \( \ce{CaCl2} \).
2. Mass of \( \ce{CaCl2} \) = 0.015 x 111 = 1.665 g.
3. Percentage = 1.665/2.22 x 100 = 75%.

Teacher's Note:
1. Conserve calcium: moles of Ca in \( \ce{CaCl2} \) equal moles in CaO.

Answer: (c) 63%

Explanation:
1. Moles of \( \ce{CO2} \) = 0.66/44 = 0.015 mol.
2. \( \ce{2NaHCO3 -> Na2CO3 + H2O + CO2} \), so \( \ce{NaHCO3} \) = 0.03 mol = 2.52 g.
3. Percentage = 2.52/4 x 100 = 63%.

Teacher's Note:
1. Two moles of bicarbonate give one mole of carbon dioxide, so double the moles of gas.