The freezing point of a 0.1 M solution of glucose is −0.186°C. If an equal volume of 0.3 M glucose solution is added to it, the freezing point of the mixture will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → −0.372°C
Depression of freezing point is a colligative property: for a non-electrolyte such as glucose it is proportional to the concentration of the solute (for a dilute aqueous solution, molarity and molality are almost equal):
\( \displaystyle \Delta T_f=K_f\,m \)
Concentration of the mixture: equal volumes V of 0.1 M and 0.3 M glucose hold 0.1V + 0.3V = 0.4V mol of glucose in a total volume 2V.
\( \displaystyle C_{\mathrm{mix}}=\frac{0.4V}{2V}=0.2\,\mathrm{M} \)
Ratio with the first solution: Kf is the same, so ΔTf grows in the same ratio as the concentration.
\( \displaystyle \frac{\Delta T_f(\mathrm{mix})}{0.186}=\frac{0.2}{0.1} \)
\( \displaystyle \Delta T_f(\mathrm{mix})=2\times0.186=0.372\,\mathrm{K} \)
Freezing point: water freezes at 0°C, so the mixture freezes at 0 − 0.372 = −0.372°C.