Q. No.Given below are two statements:
| Statement I: | The formula of cryoscopic constant is given as \(K_f=\frac{M R T_f^2}{1000 \times \Delta H_{\text {fusion }}}.\). |
| Statement II: | \(\mathrm{K}_{\mathrm{f}}\) of water is greater than that of benzene. |
| 1. | Both Statement I and Statement II are correct. |
| 2. | Statement I is correct but Statement II is incorrect. |
| 3. | Statement I is incorrect but Statement II is correct. |
| 4. | Both Statement I and Statement II are incorrect.. |
Subtopic: Depression of Freezing Point |
Level 3: 35%-60%
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An aqueous solution of \(\mathrm{0.1 ~M}\) \(\mathrm{HA}\) shows depression in freezing point of \(0.2^{\circ} \mathrm{C}.\) If \(\mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) and assuming molarity \(=\) molality, find the dissociation constant of \(\mathrm{HA}.\)
1. \(4.50 \times 10^{-5}\)
2. \(6.25 \times 10^{-3}\)
3. \(5.625 \times 10^{-4} \)
4. \(2.65 \times 10^{-4}\)
1. \(4.50 \times 10^{-5}\)
2. \(6.25 \times 10^{-3}\)
3. \(5.625 \times 10^{-4} \)
4. \(2.65 \times 10^{-4}\)
Subtopic: Depression of Freezing Point |
64%
Level 2: 60%+
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3 g of acetic acid is dissolved in 500 g of water. The depression in the freezing point of the solution is given as \(\mathrm{x} \times 10^{-1} \mathrm{~K}.\) Calculate the value of x to the nearest integer.
[Given: \(\mathrm{K}_{\mathrm{a}} \text { of } \mathrm{CH}_3 \mathrm{COOH}=1.8 \times 10^{-5}\) and \(K_f\) of water \(=1.86 \mathrm{~K} / \mathrm{molal}\)
Density of water \(=1 \mathrm{~g} / \mathrm{mL}\) ]
[Given: \(\mathrm{K}_{\mathrm{a}} \text { of } \mathrm{CH}_3 \mathrm{COOH}=1.8 \times 10^{-5}\) and \(K_f\) of water \(=1.86 \mathrm{~K} / \mathrm{molal}\)
Density of water \(=1 \mathrm{~g} / \mathrm{mL}\) ]
| 1. | 4 | 2. | 6 |
| 3. | 2 | 4. | 1 |
Subtopic: Depression of Freezing Point |
67%
Level 2: 60%+
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What happens to freezing point of benzene, when small amount of napthalene is added to benzene?
1. Increases
2. Decreases
3. Remains unchanged
4. First decreases and then increases
1. Increases
2. Decreases
3. Remains unchanged
4. First decreases and then increases
Subtopic: Depression of Freezing Point |
77%
Level 2: 60%+
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| Statement I: | The freezing point of a solution decreases with a decrease in the amount of non-volatile solute. |
| Statement II: | The freezing point of the solution is less than that of the solvent. |
| 1. | Both Statement I and Statement II are correct. |
| 2. | Statement I is correct and Statement II is incorrect. |
| 3. | Statement I is incorrect and Statement II is correct. |
| 4. | Both Statement I and Statement II are incorrect. |
Subtopic: Depression of Freezing Point |
68%
Level 2: 60%+
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A solution containing 2 g of a non-electrolyte solute dissolved in 20 g of water has a boiling point of 373.52 K. Calculate the molecular mass of the solute.
(Given that the ebullioscopic constant (Kb) is 0.52 K·kg/mol)
1. 140 g/mol
2. 80 g/mol
3. 120 g/mol
4. 100 g/mol
(Given that the ebullioscopic constant (Kb) is 0.52 K·kg/mol)
1. 140 g/mol
2. 80 g/mol
3. 120 g/mol
4. 100 g/mol
Subtopic: Depression of Freezing Point |
70%
Level 2: 60%+
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The depression in freezing point observed for a formic acid solution of concentration \(0.5 \mathrm{~mL} \mathrm{~L}^{-1}\) is \(\mathrm{0.0405^\circ C.}\) Density of formic acid is \(1.05 \mathrm{~g} \mathrm{~mL}^{-1}\) . The Van’t Hoff factor of the formic acid solution is nearly: \(\text {(Given: for water, } \mathrm{k}_{\mathrm{f}}=1.86 \mathrm{k} \mathrm{kg} \mathrm{mol}^{-1} \text { ) }\)
1. \(0.8\)
2. \(1.1\)
3. \(1.9\)
4. \(2.4\)
1. \(0.8\)
2. \(1.1\)
3. \(1.9\)
4. \(2.4\)
Subtopic: Depression of Freezing Point |
67%
Level 2: 60%+
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Two solutions A and B are prepared by dissolving \(\mathrm{1~g}\) of non-volatile solutes X and Y, respectively in \(\mathrm{1~kg}\) of water. The ratio of depression in freezing points for A and B is found to be \(1:4.\) The ratio of molar masses of X and Y is:
1. \(1:4\)
2. \(1:0.25\)
3. \(1:0.20\)
4. \(1:5\)
1. \(1:4\)
2. \(1:0.25\)
3. \(1:0.20\)
4. \(1:5\)
Subtopic: Depression of Freezing Point |
86%
Level 1: 80%+
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Calculate the van’t Hoff factor of solute A when 0.7 g of A is dissolved in 42.0 g of water and the freezing point decreases by 0.20°C.
Given:
Molar mass of A = 93 g mol⁻¹
Kf of water = 1.86 K kg mol⁻¹
1. 0.5
2. 0.6
3. 0.7
4. 0.8
Subtopic: Depression of Freezing Point |
87%
Level 1: 80%+
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150 g of acetic acid was contaminated with 10.2 g ascorbic acid \(\left(\mathrm{C}_6 \mathrm{H}_8 \mathrm{O}_6\right)\) to lower down its freezing point by \(\left(x \times 10^{-1}\right)^{\circ} \mathrm{C} .\) The value of x: (Nearest integer)
(Given : \(\mathrm{K}_{\mathrm{f}}=3.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1};\) molar mass of ascorbic acid \(=176 \mathrm{~g} \mathrm{~mol}^{-1}\))
1. 14
2. 15
3. 16
4. 18
(Given : \(\mathrm{K}_{\mathrm{f}}=3.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1};\) molar mass of ascorbic acid \(=176 \mathrm{~g} \mathrm{~mol}^{-1}\))
1. 14
2. 15
3. 16
4. 18
Subtopic: Depression of Freezing Point |
85%
Level 1: 80%+
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