For an aqueous solution, freezing point is -0.186oC. Elevation of the boiling point of the same solution is:
(Given: Kf =1.86 K kg mol-1 and Kb = 0.512 K kg mol-1)
1. 0.186
2. 0.0512
3. 1.86
4. 5.12
| 1. | 7 | 2. | 3 |
| 3. | 1 | 4. | 2 |
The depression in freezing point of 0.01 M aqueous solutions of urea, sodium chloride and sodium sulphate is in the ratio of:
1. 1 : 1 : 1
2. 1 : 2 : 3
3. 1 : 2 : 4
4. 2 : 2 : 3
What is the molality of the solution of a certain solute in a solvent if there is a freezing point depression of 0.184° and if the freezing point constant is 18.4 K kg
1. 0.01
2. 1.00
3. 0.001
4. 100

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1.00 g of a non-electrolyte solute (molar mass 250g mol) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol-1, the freezing point of benzene will be lowered by:
| 1. | 0.4 K | 2. | 0.3 K |
| 3. | 0.5 K | 4. | 0.2 K |

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If the molality of the dilute solution is doubled, then the value of a molal depression constant (Kf) will be:
1. Halved
2. Tripled
3. Unchanged
4. Doubled
| 1. | \(242.8 \mathrm{~g} \mathrm{~mol}^{-1}\) | 2. | \(238.2 \mathrm{~g} \mathrm{~mol}^{-1}\) |
| 3. | \(241.8 \mathrm{~g} \mathrm{~mol}^{-1}\) | 4. | \(240.0 \mathrm{~g} \mathrm{~mol}^{-1}\) |

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A solution of sucrose (molar mass = 342 g mol–1) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be:
(kf for water = 1.86 K kg mol–1)
1. –0.372 oC
2. –0.520 oC
3. +0.372 oC
4. –0.570 oC

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The molar freezing point constant for water is 1.86°C/mole. If 342 g of cane sugar is dissolved in 1000 g of water, the solution will freeze at
1. –1.86°C
2. 1.86°C
3. –3.92°C
4. 2.42°C

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