JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 13

Answer: (d) 12 tons

Explanation:
1. Sulphur = 1.2% of 500 = 6 tons.
2. \( \ce{S + O2 -> SO2} \): 32 parts S give 64 parts \( \ce{SO2} \).
3. \( \ce{SO2} \) = 12 tons.

Teacher's Note:
1. The mass of \( \ce{SO2} \) is double the mass of S because the molar masses are 64 and 32.

Answer: (a) 39.2%

Explanation:
1. \( \ce{2KClO3 -> 2KCl + 3O2} \): 245 g of \( \ce{KClO3} \) loses 96 g of oxygen.
2. Loss = 96/245 x 100 = 39.2%.

Teacher's Note:
1. Balance the equation first. Loss is the mass of oxygen lost compared with the initial mass.

Answer: (a) 4.48 L

Explanation:
1. Pure \( \ce{CaCO3} \) = 0.80 x 25 = 20 g = 0.2 mol.
2. \( \ce{CO2} \) = 0.2 mol.
3. Volume at STP (1 atm, 273 K, 22.4 L/mol) = 0.2 x 22.4 = 4.48 L.

Teacher's Note:
1. Use only the pure part of the sample in the calculation.

Answer: (c) 20 g

Explanation:
1. Moles of \( \ce{CHCl3} \) = 23.9/119.5 = 0.2 mol.
2. Theoretical acetone needed = 0.2 mol (1 : 1 ratio).
3. At 58% yield: 0.2/0.58 = 0.345 mol x 58 g/mol = 20 g.

Teacher's Note:
1. Divide by the fractional yield when working backwards from the product.

Answer: (d) 50%

Explanation:
1. \( \ce{2KHCO3 -> K2CO3 + H2O + CO2} \). Moles of \( \ce{KHCO3} \) = 200/100 = 2 mol.
2. Theoretical \( \ce{CO2} \) = 1 mol = 44 g.
3. Yield = 22/44 x 100 = 50%.

Teacher's Note:
1. Percentage yield = actual / theoretical x 100.

Answer: (a) 31.4%

Explanation:
1. \( \ce{N2O4 <=> 2NO2} \). Molar mass of the mixture = 2 x 35 = 70 g/mol.
2. Normal molar mass = 92. 92/70 = 1 + alpha.
3. alpha = 0.314, which is 31.4%.

Teacher's Note:
1. For dissociation, theoretical molar mass divided by observed molar mass equals 1 + alpha (n - 1).

Answer: (c) 64 g/mol

Explanation:
1. Start with 1 mol \( \ce{SO3} \): 0.5 mol remains, 0.5 mol \( \ce{SO2} \) and 0.25 mol \( \ce{O2} \) form.
2. Total moles = 1.25. Total mass = 80 g.
3. Average molar mass = 80/1.25 = 64 g/mol.

Teacher's Note:
1. Mass is conserved but total moles increase, so the molar mass falls.

Answer: (c) 40%

Explanation:
1. Molar mass of mixture = 20 x 2 = 40 g/mol.
2. For 1 mol of mixture: 32n + 48(1 - n) = 40, so n = 0.5 mol \( \ce{O2} \).
3. Mass of \( \ce{O2} \) = 16 g out of 40 g = 40%.

Teacher's Note:
1. Density ratio at the same conditions equals molar mass ratio. The answer asks for mass percent, not mole percent.

Answer: (c) 4 mol

Explanation:
1. \( \ce{Br2 + Cl2 -> 2BrCl} \) and \( \ce{Br2 + 3Cl2 -> 2BrCl3} \).
2. \( \ce{BrCl} \) : \( \ce{BrCl3} \) = 5 : 1 means \( \ce{Br2} \) used in the two reactions is 2.5 mol and 0.5 mol.
3. \( \ce{Cl2} \) = 2.5 x 1 + 0.5 x 3 = 4 mol.

Teacher's Note:
1. Find how much bromine goes into each product, then add the chlorine used in each parallel reaction.

Answer: (a) 640 g

Explanation:
1. 1 mol \( S_8 \) gives 8 mol \( \ce{SO2} \).
2. 8 mol \( \ce{SO2} \) gives 8 mol \( \ce{SO3} \) (2 : 2 ratio).
3. Mass = 8 x 80 = 640 g.

Teacher's Note:
1. In a sequence, multiply the mole ratios step by step. Here each S atom becomes one \( \ce{SO3} \).

Answer: (a) 62 g

Explanation:
1. Moles of \( \ce{KO2} \) = 100/71 = 1.408 mol.
2. \( \ce{KO2} \) gives an equal number of moles of KOH, so KOH = 1.408 mol.
3. \( \ce{CO2} \) absorbed = 1.408 mol x 44 = 62 g.

Teacher's Note:
1. The sequence has 1 : 1 mole ratios from \( \ce{KO2} \) to KOH to \( \ce{CO2} \). Multiply by the molar mass at the end.

Answer: (c) 1.2 g

Explanation:
1. Molar mass of \( \ce{Mg2P2O7} \) = 48 + 62 + 112 = 222 g/mol, so 1.11 g = 0.005 mol.
2. P atoms = 2 x 0.005 = 0.01 mol, and each \( \ce{NaH2PO4} \) has one P.
3. Mass of \( \ce{NaH2PO4} \) = 0.01 x 120 = 1.2 g.

Teacher's Note:
1. Conserve phosphorus through the steps. One \( \ce{Mg2P2O7} \) holds two P atoms.

Answer: (d) 4

Explanation:
1. Mass of Fe in 89600 g = 0.0025 x 89600 = 224 g.
2. Atoms of Fe per molecule = 224/56 = 4.

Teacher's Note:
1. Mass of the element per mole of compound divided by its atomic mass gives atoms per molecule.

Answer: (a) 2%

Explanation:
1. Molar mass of \( \ce{BaSO4} \) = 233 g/mol, so moles = 0.001.
2. Mass of S = 0.001 x 32 = 0.032 g.
3. Percentage = 0.032/1.6 x 100 = 2%.

Teacher's Note:
1. All sulphur ends up in \( \ce{BaSO4} \), one S per formula unit.

Answer: (a) 60%

Explanation:
1. Moles of AgCl = 71.75/143.5 = 0.5 mol = moles of Ag.
2. Mass of Ag = 0.5 x 108 = 54 g.
3. Percentage = 54/90 x 100 = 60%.

Teacher's Note:
1. Silver is conserved between the coin and the AgCl precipitate.

Answer: (a) 7%

Explanation:
1. n of \( \ce{N2} \) = (0.821 x 0.150)/(0.0821 x 300) = 0.005 mol.
2. Mass of N = 0.005 x 28 = 0.14 g.
3. Percentage = 0.14/2.0 x 100 = 7%.

Teacher's Note:
1. Get moles of nitrogen gas from PV = nRT and then convert to mass of nitrogen.

Answer: (b) 19

Explanation:
1. Molar mass = 240/0.1046 = 2294.
2. Mass of end group \( \text{Br}_3\text{C}_6\text{H}_2 \) = 240 + 72 + 2 = 314.
3. n = (2294 - 314)/104 = 19.

Teacher's Note:
1. Use the mass percent of bromine to find the total molar mass, then divide the remainder by the repeating unit mass.

Answer: (b) \( \ce{C6H4Cl2} \)

Explanation:
1. Moles: C = 49/12 = 4.08, H = 2.7, Cl = 48.3/35.5 = 1.36. Ratio is 3 : 2 : 1.
2. Empirical formula \( \ce{C3H2Cl} \) has mass 73.5.
3. 147/73.5 = 2, so molecular formula = \( \ce{C6H4Cl2} \).

Teacher's Note:
1. First find the empirical formula, then scale it with molar mass divided by empirical mass.

Answer: (d) 4

Explanation:
1. Moles of N atoms = 6/14 = 0.4286 mol.
2. Molar mass of the complex = 52 + 17x + 160 = 212 + 17x.
3. x x 30/(212 + 17x) = 0.4286, which gives x = 4.

Teacher's Note:
1. All nitrogen ends up in the ammine ligands, x per complex. Set up the atom balance.

Answer: (d) 6

Explanation:
1. Mass of H in 58 g = 0.104 x 58 = 6.03 g.
2. H atoms = 6 (the compound is \( \ce{C3H6O} \)).

Teacher's Note:
1. Mass of the element in one mole divided by its atomic mass gives atoms per molecule.

Answer: (d) 120 g/mol

Explanation:
1. Moles of Pt = 0.24/195 = 1.231 x \( 10^{-3} \) mol, so moles of salt are the same.
2. Molar mass of the salt = 0.80/1.231 x \( 10^{-3} \) = 650 g/mol.
3. Salt mass = 2(B + 1) + 195 + 213 = 2B + 410 = 650, so B = 120.

Teacher's Note:
1. Each salt formula unit contains two base molecules and one platinum atom. Work from platinum to the salt's molar mass.