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%
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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

Subtopic:  Depression of Freezing Point |
 88%
From NCERT
NEET - 2020
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Freezing point of an aqueous solution is -0.166°C. 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 |
 86%
From NCERT
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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 |
 82%
From NCERT
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