Question

Mole fraction of a non electrolyte in aqueous solution is 0.07. If $K_f$ is ${1.86}^{0}mo{l}^{-1}Kg$, depression in freezing point $\Delta T_f$ is:

• A. ${0.26}^0$
• B. ${1.866}^0$
• C. ${0.13}^0$
• D. ${7.78}^0$

Answer 1

The correct option is D ${7.78}^0$

Given:

Mole fraction of a non electrolyte in aqueous solution = 0.07

$K_f=1.86$

Depression in freezing point = ?

Solution :

Mole fraction of solvent = 1- mole fraction of non electrolyte

$\Rightarrow X_1=1-X_2=1-0.07=0.93$

Depression in freezing point,

$\Delta T_f=i\times m\times K_f$

Since the solute is non electrolyte, i=1

Molarity, m= $\frac{No.Ofmolesofsolute}{MassofsolventinKg}$

$\Rightarrow m=\frac{0.07}{0.93\times 18}$

$\Rightarrow m=4.18m$

Therefore, $\Delta T_f=1\times 4.18\times 1.86$ = 7.78 °C

Hence, correct option is D.