JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 14

Answer: (c) 153 g/mol

Explanation:
1. The salt is \( \text{Ag}_2\text{A} \). Its molar mass is M + 214 (two H replaced by two Ag).
2. The salt formed per 216 g of Ag is M + 214 g, so slope = (M + 214)/216 = 1.7.
3. M = 367.2 - 214 = 153 g/mol.

Teacher's Note:
1. The slope equals mass of salt per unit mass of silver. Two silver atoms replace the two hydrogens of a dibasic acid.

Answer: (c) 14.2 g

Explanation:
1. Mass of solution = 128 g, so HCl = 0.365 x 128 = 46.72 g = 1.28 mol.
2. 16 HCl gives 5 \( \ce{Cl2} \); 6 \( \ce{Cl2} \) gives 1 \( \ce{Ca(ClO3)2} \); 1 \( \ce{Ca(ClO3)2} \) gives 2 \( \ce{NaClO3} \).
3. \( \ce{NaClO3} \) = 1.28 x (5/16) x (1/6) x 2 = 0.1333 mol.
4. Mass = 0.1333 x 106.5 = 14.2 g.

Teacher's Note:
1. Chain the mole ratios step by step. Convert the volume of acid to mass using density and then to moles using the mass percent.

Answer: (d) 10.1 : 0.66 : 1.0

Explanation:
1. Per 100 g: N = 30/14 = 2.143 mol.
2. P = 2 x 10/142 = 0.1408 mol. K = 2 x 10/94 = 0.2128 mol.
3. Dividing by K: N = 10.1 and P = 0.66, so the ratio is 10.1 : 0.66 : 1.0.

Teacher's Note:
1. Convert oxide percentages to atoms of the element, remembering two P or K atoms per oxide formula unit.

Answer: (a) 77.8%

Explanation:
1. Moles of CaO = 1.12/56 = 0.02 mol = moles of \( \ce{CaCl2} \).
2. Mass of \( \ce{CaCl2} \) = 0.02 x 111 = 2.22 g.
3. NaCl = 10 - 2.22 = 7.78 g = 77.8%.

Teacher's Note:
1. Find the component that is converted, then take the remainder for the other.

Answer: (b) \( \ce{Fe2O3 + 2Al -> Al2O3 + 2Fe} \)

Explanation:
1. Balance Fe: 2 Fe on the left, so 2 Fe on the right.
2. Balance Al: 2 Al on the right, so 2 Al on the left.
3. Check O: 3 on each side.

Teacher's Note:
1. Balance metals first and check oxygen last.

Answer: (b) 80 : 27

Explanation:
1. Stoichiometric amounts react completely: 1 mol \( \ce{Fe2O3} \) (160 g) with 2 mol Al (54 g).
2. Any excess of one reactant adds mass without adding energy, so the stoichiometric ratio is best.
3. 160 : 54 = 80 : 27.

Teacher's Note:
1. Energy per unit mass is greatest when no reactant is wasted, which is the stoichiometric ratio.

Answer: (a) 10000 units

Explanation:
1. \( \ce{Fe2O3} \) = 16000/160 = 100 mol. Al = 2700/27 = 100 mol.
2. Al needed per mol \( \ce{Fe2O3} \) is 2, so Al is limiting: only 50 mol \( \ce{Fe2O3} \) reacts.
3. Energy = 50 x 200 = 10000 units.

Teacher's Note:
1. Find the limiting reagent before using the energy per mole.

Answer: (a) 59.7%

Explanation:
1. \( \ce{KClO3} \) = 1/122.5 = 0.008163 mol. \( \ce{O2} \) = 0.112/22.4 = 0.005 mol.
2. Reaction 1 uses x mol with 1.5x = 0.005, so x = 0.003333. Remaining 0.004830 mol goes by reaction 2.
3. \( \ce{KClO4} \) = 0.75 x 0.004830 = 0.003623 mol = 0.5017 g.
4. Residue mass = 1 - 0.005 x 32 = 0.84 g. Percentage = 0.5017/0.84 = 59.7%.

Teacher's Note:
1. Use the oxygen evolved to split the sample between the two reactions. Remember that the residue excludes the oxygen that escaped.

Answer: (a) 5.74 mg

Explanation:
1. Molar mass = 52 + 90 + 35.5 + 71 + 18 = 266.5 g/mol, so 5.33 mg = 0.02 mmol.
2. Only the two ionic chloride ions precipitate; the Cl inside the complex ion does not.
3. AgCl = 0.04 mmol x 143.5 = 5.74 mg.

Teacher's Note:
1. Only chloride outside the coordination sphere reacts with silver nitrate.

Answer: (d) \( 6.0\times10^{19} \)

Explanation:
1. The carbonate is \( M_2CO_3 \), with 3 O atoms. Molar mass = 48/0.48 = 100 g/mol.
2. Moles = 0.005/100 = 5 x \( 10^{-5} \) mol.
3. Metal atoms = 2 x 5 x \( 10^{-5} \) x \( 6.0\times10^{23} \) = \( 6.0\times10^{19} \).

Teacher's Note:
1. Use the oxygen percent to get the molar mass, then remember two metal atoms per formula unit.

Answer: (d) \( A_3B_2 \)

Explanation:
1. AO has molar mass 40, so 3 g = 0.075 mol, so A = 0.075 mol (1.8 g).
2. B = 2.5 - 1.8 = 0.7 g = 0.05 mol.
3. A : B = 0.075 : 0.05 = 3 : 2, so the formula is \( A_3B_2 \).

Teacher's Note:
1. Conserve A from the product oxide. The rest of the compound's mass is B.

Answer: (b) 222 g

Explanation:
1. Ca = \( 2.4\times10^{24}/6\times10^{23} \) = 4 mol.
2. P = 380/760 = 0.5 atm. n of \( \ce{Cl2} \) = (0.5 x 96)/(0.08 x 300) = 2 mol.
3. \( \ce{Ca + Cl2 -> CaCl2} \): \( \ce{Cl2} \) is limiting, so \( \ce{CaCl2} \) = 2 mol = 222 g.

Teacher's Note:
1. Convert the pressure to atm and the temperature to K before using PV = nRT.

Answer: (b) 11 g

Explanation:
1. \( \ce{SO2} \) 9.6 g = 0.15 mol needs 0.05 mol \( \ce{P4S3} \).
2. \( \ce{P4O10} \) 9.6 g = 0.0338 mol needs 0.0338 mol \( \ce{P4S3} \).
3. The larger requirement controls: 0.05 mol x 220 = 11 g.

Teacher's Note:
1. Work out the reactant needed for each product separately and take the larger.

Answer: (d) 5 mol

Explanation:
1. \( \ce{H4P2O7} \) = 534/178 = 3 mol. NaOH = \( 3.0\times10^{24}/6\times10^{23} \) = 5 mol.
2. 3 mol acid needs 6 mol NaOH, so NaOH is limiting.
3. \( \ce{H2O} \) = 5 mol (1 mol per mol NaOH).

Teacher's Note:
1. Compare moles divided by coefficients to find the limiting reagent first.

Answer: (b) 1.70 g

Explanation:
1. Mass of reactants = mass of products = 1.435 + 0.85 = 2.285 g.
2. Mass of \( \ce{AgNO3} \) = 2.285 - 0.585 = 1.70 g.

Teacher's Note:
1. In a closed reaction the total mass before and after is the same.

Answer: (b) 3.3 g

Explanation:
1. Total mass before = 6.3 + 15 = 21.3 g.
2. Mass after = 18 g, so the lost mass is the gas that escaped.
3. \( \ce{CO2} \) = 21.3 - 18 = 3.3 g.

Teacher's Note:
1. A gas that escapes accounts for the missing mass.

Answer: (d) 86.62%

Explanation:
1. Mass of oxide = 6.488 + 1.002 = 7.490 g.
2. Percentage of Pb = 6.488/7.490 x 100 = 86.62%.
3. This equals 86.62% from the second method, which illustrates the law of definite proportions.

Teacher's Note:
1. Find the percentage of the element in the compound from each source and compare them.

Answer: (a) 1 : 2 : 3 : 5

Explanation:
1. For 28 g (2 mol atoms) of N: N2O has 16 g of O, NO has 32 g, N2O3 has 48 g, N2O5 has 80 g.
2. Ratio = 16 : 32 : 48 : 80 = 1 : 2 : 3 : 5.
3. A simple whole number ratio confirms the law of multiple proportions.

Teacher's Note:
1. Fix the mass of one element, then compare the masses of the other.

Answer: (d) 1 : 4

Explanation:
1. Oxide 1: oxygen = 0.5 g per 0.5 g of the element, so 1 g per g.
2. Oxide 2: oxygen = 3.2 g per 0.8 g of the element, so 4 g per g.
3. Ratio = 1 : 4, a simple ratio.

Teacher's Note:
1. Find oxygen per gram of the element in each oxide and compare.

Answer: (d) 1 : 2

Explanation:
1. 56 mL of \( \ce{N2} \) = 0.0025 mol = 0.07 g of N in each case.
2. Oxygen: 0.11 - 0.07 = 0.04 g and 0.15 - 0.07 = 0.08 g.
3. Ratio of oxygen for the same 0.07 g of N = 0.04 : 0.08 = 1 : 2.

Teacher's Note:
1. Both samples contain the same mass of nitrogen, so compare the oxygen masses directly.