Question

The freezing point of 0.1 M glucose solution is $-{1.86}^{0}C$. If an equal volume of 0.3 M glucose solution is added, the freezing of the mixture will be :

• A. $-{7.44}^{0}C$
• B. $-{5.58}^{0}C$
• C. $-{3.72}^{0}C$
• D. $-{2.79}^{0}C$

Answer 1

The correct option is C $-{3.72}^{0}C$

The expression for the depression in the freezing point and the molality is $K_f=\frac{\Delta T_f}{m}$.

The molality is 0.1 m.

The freezing point of 0.1 M glucose solution is ${1.86}^{0}C$.

The depression in freezing point is $0-\left({1.86}^{0}C\right)={1.86}^{0}C$.

Substitute values in the above expression.

$K_f=\frac{1.86}{0.1}=18.6$

When an equal volume of 0.3 M glucose solution is added, the molality of the resulting solution is given by the expression

$0.1\times \vee +0.3\vee =M_3\times 2\vee$.

Hence, $M_3=0.2M$.

The expression for the freezing point depression will be $\Delta T_f=18.6\times 0.2={3.72}^{o}C$.

The freezing point of the mixture will be $T_f=0-3.72\simeq -{3.72}^{o}C$.

Note: Molarity $\approx$ molality.

Option C is correct.