The freezing point 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) :

Question

2g of benzoic acid (C₆H₅COOH) dissolved in 25g of benzene lowers its freezing point by 1.62 K. Molal depression constant is 4.9 K kg/mol. The value of ‘i’ when benzoic acid dimersises would be :

Answer 1

To solve this problem, we will use the formula for freezing point depression:

\[\Delta T_f = K_f \times m\]

where \(\Delta T_f\) is the freezing point depression, \(K_f\) is the freezing point depression constant, and \(m\) is the molality of the solution.

Given:

Freezing point depression constant, \(K_f = 5.12\ K\ kg\ mol^{-1}\)
Molality of the solution, \(m = 0.078\ m\)

Substituting the given values into the formula:

\[\Delta T_f = 5.12\ K\ kg\ mol^{-1} \times 0.078\ mol\ kg^{-1}\]

Calculating further:

\[\Delta T_f = 0.39936\ K\]

Rounding off to two decimal places, we get:

\[\Delta T_f = 0.40\ K\]

Thus, the freezing point depression for the solution is 0.40 K.

This matches the given correct answer option:

0.40 K

Answer 2