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Q. No.Freezing point of an aqueous solution is -0.166C. Elevation of boiling point of same solution would be-
(Kb = 0.512 K m-1 and Kf = 1.66 K m-1)
| 1. | 0.18°C | 2. | 0.05°C |
| 3. | 0.09°C | 4. | 0.23°C |
The freezing point of depression constant (Kf ) of benzene is 5.12 K kg mol–1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is:
(rounded off up to two decimal places)
1. 0.80 K
2. 0.40 K
3. 0.60 K
4. 0.20 K
A 0.1 molal aqueous solution of a weak acid (HA) is 30 % ionized. If Kf for water is 1.86 °C/m, the freezing point of the solution will be:
| 1. | –0.24 °C | 2. | –0.18 °C |
| 3. | –0.54 °C | 4. | –0.36 °C |
1. 1.0 m KBr
2. 0.75 m C6H12O6
3. 0.5 m MgCl2
4. 0.25 m Ga2(SO4)3
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
Choose a false statement among the following:
| 1. | Raoult's law states that the vapour pressure of a component over a solution is proportional to its mole fraction. |
| 2. | The osmotic pressure (\(\pi\)) of a solution is given by the equation \(\pi\)=MRT, where M is the molarity of the solution. |
| 3. | The correct order of osmotic pressure for 0.10 M aqueous solution of each compound is BaCl2 > KCl > CH3COOH > Sucrose. |
| 4. | Two sucrose solutions of the same molarity prepared in different solvents will have the same depression in the freezing point. |
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