- 1(current)
Q. No.Match the laws/colligative properties given in Column-I with expressions given in Column-II.
| Column-I | Column-II | ||
| A. | Raoult’s law | I. | \(\mathrm{\pi=C R T}\) |
| B. | Osmotic pressure | II. | \(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{~m}\) |
| C. | Elevation of boiling point | III. | \(\mathrm{p=x_1 p_1^0+x_2 p_2^0}\) |
| D. | Depression in freezing point | IV. | \(\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{~m}\) |
Codes:
| A | B | C | D | |
| 1. | I | III | IV | I |
| 2. | I | II | III | IV |
| 3. | I | IV | III | II |
| 4. | III | I | IV | II |
Subtopic: Elevation of Boiling Point | Depression of Freezing Point |
94%
Level 1: 80%+
Hints
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:
1. 0.80 K
2. 0.40 K
3. 0.60 K
4. 0.20 K
Subtopic: Depression of Freezing Point |
88%
Level 1: 80%+
NEET - 2020
Hints
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 |
Subtopic: Depression of Freezing Point |
87%
Level 1: 80%+
Hints
Links
In comparison to a 0.01 M solution of glucose, the depression in the freezing point of a 0.01 M MgCl2 solution will be:
| 1. | Same | 2. | About twice |
| 3. | About three times | 4. | About six times |
Subtopic: Depression of Freezing Point |
83%
Level 1: 80%+
Hints
Links
- 1(current)
Q. No.
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