JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 19

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. The mass of one molecule in amu is numerically equal to the mass of one mole in grams, because 1 amu = 1/\( N_A \) g.
2. The Reason states exactly this molar relationship, so it explains the Assertion.

Teacher's Note:
1. The numerical equality of amu per molecule and g per mole follows from the definition of the mole.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. In dilute aqueous solution, 1 kg of solvent is about 1 L of solution.
2. This happens because the density of water is about 1 g/\( cm^3 \).

Teacher's Note:
1. Molarity and molality are nearly equal when the solution is dilute and its density is close to 1.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. Molarity uses volume, which is easy to measure with a flask or pipette.
2. This explains the common use of molarity.

Teacher's Note:
1. Practical convenience favours volume-based concentration units.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. Both depend only on masses or moles, which do not change with temperature.
2. Volume changes with temperature, but neither unit involves volume.

Teacher's Note:
1. Only volume-based units such as molarity change with temperature.

Answer: (d) 30.77% and 18.70%

Explanation:
1. \( Ti_{0.75}O \): mass = 0.75 x 48 + 16 = 52 g; O = 16/52 = 30.77%.
2. \( TiO_{0.69} \): mass = 48 + 0.69 x 16 = 59.04 g; O = 11.04/59.04 = 18.70%.

Teacher's Note:
1. Oxygen percentage is highest where the least titanium is present per oxygen.

Answer: (b) \( 7.09\times10^{7} \)

Explanation:
1. Volume = \( V = \pi r^2 l = 3.14\times(75\times10^{-8})^2\times5000\times10^{-8} = 8.84\times10^{-17} \) \( cm^3 \).
2. Mass of one particle = \( 8.84\times10^{-17}/0.75 = 1.18\times10^{-16} \) g.
3. Molar mass = \( 1.18\times10^{-16}\times6.022\times10^{23} = 7.09\times10^{7} \) g/mol.

Teacher's Note:
1. Radius is half the diameter. Convert Angstrom to cm before computing the volume.

Answer: (b) 33.33%

Explanation:
1. Let chalk = x g. It loses 44% of its mass as \( \ce{CO2} \), that is 0.44x.
2. Clay = (5 - x) g loses 0.11(5 - x).
3. 0.44x + 0.11(5 - x) = 1.1, so 0.33x = 0.55 and x = 1.667 g.
4. Percentage = 1.667/5 x 100 = 33.33%.

Teacher's Note:
1. Write the total mass loss as the sum of the two separate losses.

Answer: (b) K4Fe(CN)6; 6 mol CO and 2 mol \( \mathrm{(NH_4)_2SO_4} \)

Explanation:
1. Step (i): 1 mol of the complex needs 3 mol \( \ce{H2SO4} \), so K4Fe(CN)6 (1 vs 5/3) is limiting. It gives 6 mol HCN, then 6 mol HCOOH and 6 mol \( \ce{NH3} \).
2. \( \ce{H2SO4} \) left = 5 - 3 = 2 mol, which converts 4 mol \( \ce{NH3} \) into 2 mol \( \mathrm{(NH_4)_2SO_4} \).
3. HCOOH gives 6 mol CO.

Teacher's Note:
1. Track how much sulphuric acid is left after step (i); it decides how much ammonium sulphate forms.

Answer: (d) 0.5 mol \( A_2B_4 \) and 0.5 mol \( A_3B_4 \)

Explanation:
1. 2 mol \( B_2 \) makes 1 mol \( A_2B_4 \) using 1 mol \( A_2 \). 0.25 mol \( A_2 \) remains.
2. 0.25 mol \( A_2 \) converts 0.5 mol \( A_2B_4 \) into 0.5 mol \( A_3B_4 \).
3. Final: 0.5 mol \( A_2B_4 \) and 0.5 mol \( A_3B_4 \).

Teacher's Note:
1. Fill the first reaction using all the B2, then use leftover A2 to convert \( A_2B_4 \) into \( A_3B_4 \).

Answer: (a) 1.1 mol

Explanation:
1. Let x mol AB and y mol \( AB_4 \) form: x + y = 1 and 100x + 200y = 140.
2. y = 0.4 and x = 0.6.
3. \( B_2 \) used = 0.6 x 0.5 + 0.4 x 2 = 0.3 + 0.8 = 1.1 mol.

Teacher's Note:
1. Use one equation for the amount of A and one for the heat to find the split.

Answer: (d) 1.14 g

Explanation:
1. Moles of \( \ce{Mg2P2O7} \) = 1.054/222 = 4.75 x \( 10^{-3} \) mol.
2. P atoms = 9.50 x \( 10^{-3} \) mol, so \( \ce{NaH2PO4} \) = 9.50 x \( 10^{-3} \) mol.
3. Mass = 9.50 x \( 10^{-3} \) x 120 = 1.14 g.

Teacher's Note:
1. Conserve phosphorus: two P atoms per \( \ce{Mg2P2O7} \).

Answer: (d) About 8 hours

Explanation:
1. To gain 1 g of starch (162 g per 6 C), carbon needed = 6/162 = 0.0370 mol of \( CO_2 \).
2. Time = 0.0370/\( 4.7\times10^{-3} \) = 7.9 hours, about 8 hours.

Teacher's Note:
1. Convert 1 g of starch to moles of carbon fixed, then divide by the hourly rate.

Answer: (b) 4

Explanation:
1. \( \ce{Cl2} \) = 5 mol. Step 1: 60% reacts = 3 mol, giving 3 mol KCl and 3 mol KClO.
2. Step 2: \( \ce{3KClO -> 2KCl + KClO3} \); 50% of 3 mol = 1.5 mol KClO reacts, giving 1 mol KCl and 0.5 mol \( \ce{KClO3} \).
3. Step 3: \( \ce{4KClO3 -> 3KClO4 + KCl} \); 80% of 0.5 mol = 0.4 mol reacts, giving 0.1 mol KCl.
4. Total KCl = 3 + 1 + 0.1 = 4.1, which rounds to 4.

Teacher's Note:
1. Apply the percentage yield at each step and add up the KCl from all steps.

Answer: (a) 62%

Explanation:
1. Active material = 95.5 g. Let \( \ce{Cu2S} \) = x g (159 g/mol, 127 g Cu), CuS = (95.5 - x) g (95.5 g/mol, 63.5 g Cu).
2. (127/159)x + (63.5/95.5)(95.5 - x) = 71.8.
3. 0.7987x + 63.5 - 0.6649x = 71.8, so x = 62.0 g, which is 62%.

Teacher's Note:
1. Write the copper balance for the two minerals and solve for the one unknown.

Answer: (b) 29.2 mL

Explanation:
1. Molarity of the stock = 10 x 1.25 x 50/36.5 = 17.1 M.
2. M1 V1 = M2 V2: V1 = 0.5 x 1000/17.1 = 29.2 mL.

Teacher's Note:
1. Find the stock molarity from density and percentage, then use dilution.

Answer: (b) 70 mL

Explanation:
1. NaOH = 4 mmol neutralises 2 mmol \( \ce{H2SO4} \).
2. Acid left = 0.2V - 2 mmol in (V + 40) mL.
3. (0.2V - 2)/(V + 40) = 6/55, so 11V - 110 = 6V + 240 and V = 70 mL.

Teacher's Note:
1. Subtract the acid neutralised by the base, then equate the remaining concentration to the given value.

Answer: (a) 0.2 \( cm^2 \)

Explanation:
1. HCl = 0.05 x 12 = 0.6 mmol. \( \ce{2Al + 6HCl -> 2AlCl3 + 3H2} \), so Al = 0.2 mmol = 5.4 mg.
2. Volume of Al = 5.4 x \( 10^{-3} \)/2.70 = \( 2\times10^{-3} \) \( cm^3 \).
3. Area = volume/thickness = \( 2\times10^{-3}/0.01 = 0.2 \) \( cm^2 \).

Teacher's Note:
1. Convert thickness to cm (0.10 mm = 0.01 cm). Area = volume / thickness.

Answer: (c) 1.06

Explanation:
1. 12 H atoms per formula unit, so 3.18/12 = 0.265 mol of the compound.
2. O atoms = 4 x 0.265 = 1.06 mol.

Teacher's Note:
1. Moles of the compound link the moles of the two kinds of atoms.

Answer: (a) 10% less

Explanation:
1. Molar mass of \( \ce{H2O} \) = 18; of \( \ce{D2O} \) = 20.
2. (20 - 18)/20 = 10% less.

Teacher's Note:
1. Heavy water has two deuterium atoms of mass 2 each.

Answer: (c) 71.04

Explanation:
1. At the same conditions, mass ratio = molar mass ratio.
2. Molar mass = 2.22 x 32 = 71.04 g/mol.

Teacher's Note:
1. For equal volumes of gases, the ratio of masses equals the ratio of molar masses.

Answer: (a) 0.5 : 0.1 : 0.4

Explanation:
1. Test option (a): 0.5(28) + 0.1(46) + 0.4(92) = 14 + 4.6 + 36.8 = 55.4. Correct.
2. After complete dissociation: moles = 0.5 + 0.1 + 0.8 = 1.4 and mass = 55.4 g, so mean molar mass = 39.6.
3. The other options give a different mean molar mass.

Teacher's Note:
1. Use mass conservation: total mass stays constant while moles change on dissociation.

Answer: (d) 1 : 2 : 3

Explanation:
1. Mass of the element combining with 16 g of O: 42.7 g, 21.3 g and 14.2 g (the equivalent masses).
2. Valency is inversely proportional to the equivalent mass: 1/42.7 : 1/21.3 : 1/14.2 = 1 : 2 : 3.

Teacher's Note:
1. Valency = atomic mass / equivalent mass, and equivalent mass is the mass that combines with 8 g of oxygen.

Answer: (a) 916.2 g

Explanation:
1. \( \ce{C12H22O11 + 12O2 -> 12CO2 + 11H2O} \). Sucrose = 34/342 = 0.0994 mol per hour.
2. \( \ce{O2} \) per hour = 12 x 0.0994 = 1.193 mol = 38.2 g.
3. For 24 hours: 38.2 x 24 = 916.2 g.

Teacher's Note:
1. Compute oxygen per hour, then multiply by 24.