The freezing point of 0.1 M solution of glucose is –1.86°C. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will be:

–3.72°C

The correct answer is option 3. –3.72°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 \(1.86^oC\)

The depression in the freezing point is \(0 − (1.86^oC) = 1.86^oC\)

Substituting the values in the above expression

\(K_f = \frac{1.86}{0.1} = 18.6\)'

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

\(0.1V + 0.3V = M_3 × 2V\)

\(∴ M_3 = 0.2M\)

The expression for the freezing point depression will be

\(T_f = 0 − 3.72 ≅ −3.72ºC\)