JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 11

Answer: (a) \( \frac{6}{7} \)

Explanation:
1. Mixture molar mass = 2.5 x 22.4 = 56.
2. \( SO_3 \to SO_2 + \tfrac12 O_2 \), so moles become 1 + a/2.
3. 80/(1 + a/2) = 56 gives a/2 = 3/7, so a = 6/7.

Teacher's Note:
1. Density at STP times 22.4 gives molar mass of the mixture.

Answer: (d) 80%

Explanation:
1. Aniline 9.30/93 = 0.1 mol, so theoretical iodobenzene = 0.1 mol x 204 = 20.4 g.
2. Yield = 16.32/20.4 x 100 = 80%.

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

Answer: (b) 8.75 kg

Explanation:
1. 20 kg \( CaC_2 \) = 20000/64 = 312.5 mol.
2. Each mole gives one mole of \( C_2H_4 \) (28 g).
3. Polymer = 312.5 x 28 = 8750 g = 8.75 kg.

Teacher's Note:
1. Follow one mole through the chain; the ratio is 1 : 1 at each stage.

Answer: (a) 1.9 moles

Explanation:
1. 128 g \( SO_2 \) = 2 mol.
2. The three steps are 1 : 1 : 1, so 2 mol \( CaSO_4 \) at 100%.
3. At 95% efficiency: 1.9 mol.

Teacher's Note:
1. The overall conversion is 1 mol SO2 to 1 mol CaSO4. Then apply the efficiency.

Answer: (a) 4

Explanation:
1. Take n mol \( KClO_3 \). In (i) KCl = n mol.
2. In (ii) KCl = n/4 mol.
3. Ratio = n : n/4 = 4 : 1, so x = 4.

Teacher's Note:
1. The same compound is formed, so the mole ratio equals the mass ratio.

Answer: (c) 7

Explanation:
1. Water lost = 6.3 g = 0.35 mol.
2. Anhydrous \( Na_2SO_4 \) = 13.4 - 6.3 = 7.1 g = 0.05 mol (142 g/mol).
3. x = 0.35/0.05 = 7.

Teacher's Note:
1. Moles of water divided by moles of anhydrous salt gives x.

Answer: (c) 13.6

Explanation:
1. The formula is \( C_5H_8NO_4Na \).
2. Molar mass = 60 + 8 + 14 + 64 + 23 = 169.
3. Na% = 23/169 x 100 = 13.6%.

Teacher's Note:
1. Write the full molecular formula from the structure before adding up the molar mass.

Answer: (d) HCHO, \( CH_3COOH \), \( C_6H_{12}O_6 \)

Explanation:
1. HCHO, \( CH_3COOH \) and \( C_6H_{12}O_6 \) all have the empirical formula \( CH_2O \), so each has 40% C.
2. The other series have different carbon percentages (for example \( C_{10}H_8 \) has 93.75% while \( C_2H_2 \) has 92.3%).

Teacher's Note:
1. Compounds with the same empirical formula have the same percentage composition.

Answer: (c) \( N_2O_4 \)

Explanation:
1. N = 0.305 x 92 = 28 g, which is 2 atoms.
2. O = 0.695 x 92 = 64 g, which is 4 atoms.
3. The formula is \( N_2O_4 \).

Teacher's Note:
1. Find the mass of each element in one mole, then divide by the atomic mass.

Answer: (a) \( C_3H_8 \)

Explanation:
1. By Avogadro's law equal volumes have equal moles, so equal masses means equal molar masses.
2. Molar mass = 44, and \( C_3H_8 \) = 36 + 8 = 44.

Teacher's Note:
1. Same volume and same mass means the same molar mass.

Answer: (b) \( CH_3COOH \)

Explanation:
1. Glucose \( C_6H_{12}O_6 \) has empirical formula \( CH_2O \).
2. Acetic acid \( C_2H_4O_2 \) reduces to \( CH_2O \).
3. The others reduce to \( C_2H_4O \), \( CH_4O \) and \( CH_3 \).

Teacher's Note:
1. Divide the subscripts by their highest common factor.

Answer: (c) \( X_2Y_3 \)

Explanation:
1. Mole ratio X : Y = 1/30 : 1/20 = 2 : 3.
2. The formula is \( X_2Y_3 \).

Teacher's Note:
1. Equal masses give mole ratio inversely proportional to atomic masses.

Answer: (d) 170

Explanation:
1. Moles of vapour = 67.2/22400 = 0.003.
2. M = 0.510/0.003 = 170 g/mol.

Teacher's Note:
1. The volume of air displaced equals the volume of vapour at the same conditions.

Answer: (d) \( C_5H_6O_5 \)

Explanation:
1. The dibasic acid forms \( Ag_2 \)-salt: 216 g of Ag per mole of salt.
2. 216/M = 0.6, so M(salt) = 360.
3. Acid = 360 - 216 + 2 = 146, which matches \( C_5H_6O_5 \) (60 + 6 + 80 = 146).

Teacher's Note:
1. Replace two Ag by two H to go from the salt to the acid.

Answer: (a) 9.3 g

Explanation:
1. Moles = \( 2.01\times10^{23}/6.02\times10^{23} = 0.334 \).
2. Mass = 0.334 x 28 = 9.3 g.

Teacher's Note:
1. Number divided by N_A gives moles.

Answer: (a) \( C_6H_8N_2 \)

Explanation:
1. Total parts = 13.5, so 108 g contains 72 g C, 8 g H and 28 g N.
2. Atoms: C = 6, H = 8, N = 2.
3. The formula is \( C_6H_8N_2 \).

Teacher's Note:
1. Scale the weight ratio to the molar mass, then divide by atomic masses.

Answer: (b) \( MCl_4 \)

Explanation:
1. Molar mass = 2 x 94.8 = 189.6.
2. Chlorine mass = 0.7475 x 189.6 = 141.7 g, which is 4 x 35.5.
3. The formula is \( MCl_4 \).

Teacher's Note:
1. Divide the chlorine mass in one mole by 35.5 to get the number of Cl atoms.

Answer: (d) 1 : 1 : 1

Explanation:
1. 16 g ozone has 16 g of O, which is 1 mol of O atoms.
2. 28 g CO is 1 mol, so 1 mol of O atoms.
3. 16 g \( O_2 \) has 1 mol of O atoms.
4. The ratio is 1 : 1 : 1.

Teacher's Note:
1. Mass of oxygen atoms is the same in 16 g of any pure oxygen allotrope.

Answer: (c) 7.5 kg

Explanation:
1. Hydrogen mass = 10% of 75 = 7.5 kg.
2. Each 1H atom becomes 2H, so the mass doubles.
3. The gain equals 7.5 kg.

Teacher's Note:
1. Doubling the mass of hydrogen adds exactly the original hydrogen mass.

Answer: (b) 84.3

Explanation:
1. One mole of \( M_2CO_3 \) gives one mole of \( CO_2 \).
2. Molar mass = 1/0.01186 = 84.3 g/mol.

Teacher's Note:
1. Mass divided by moles gives molar mass.

Answer: (c) \( C_2H_4O_3 \)

Explanation:
1. Mass ratio C : H = 6 : 1 gives atom ratio x : y = 6/12 : 1 = 1 : 2, so y = 2x.
2. Burning \( C_XH_Y \) needs x + y/4 = 1.5x molecules of \( O_2 \), which is 3x atoms of O.
3. Half of this gives z = 1.5x. For x = 2: y = 4 and z = 3, so \( C_2H_4O_3 \).

Teacher's Note:
1. Use the two conditions to get x : y : z = 2 : 4 : 3.