Match the laws given in Column I with expressions given in Column II.

Codes

The molality of a solution containing a certain solute, if there is a freezing point depression of 0.184 °C, is-

(K_f = 18.4)

1. 0.01 m

2. 10 m

3. 0.05 m

4. 100 m

The freezing point of depression constant (K_f ) 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 upto two decimal places):
1. 0.80 K
2. 0.40 K
3. 0.60 K
4. 0.20 K

When 20 g of naphthoic acid (C₁₁H₈O₂) is dissolved in 50 g of benzene (K_f = 1.72 K kg mol^–1),
a freezing point depression of 2 K is observed. The van’t Hoff factor (i) is –

Freezing point of an aqueous solution is -0.166$°$C. Elevation of boiling point of same solution would be-

(K_b = 0.512 K m^-1 and K_f = 1.66 K m^-1)

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  -

A solution containing 6.8 g of a non-ionic solute in 100 g of water was found to freeze at −0.93 ^oC. The freezing point depression constant of water is 1.86. The molecular weight of the solute is-

The freezing point depression constant for water is 1.86 ^oC m^-1. If 5.00 g Na₂SO₄ is dissolved in 45.0 g H₂O, the freezing point is changed by -3.82 ^oC. The van’t Hoff factor for Na₂SO₄ is:

The graph, amongst the following, that represents depression in the freezing point is:

1.
2.
3.
4.	None of the above

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:

 (K_f for water = 1.86 K kg mol^-1)

1. -0.372 $°$C

2. 0.372 $°$C

3. 0.572 $°$C

4. -0.572 $°$C