Question

A dilute solution prepared by dissolving a nonvolatile solute in one liter of water shows a depression in freezing point of 0.186 K. This solute neither dissociates nor associates in water. The boiling point of the solution in K (rounded off to three decimal places) is ________.
(Given: For pure water, $T_b = 373.15$ K; $K_f = 1.86$ K mol kg$^{-1}$; $K_b = 0.51$ K mol kg$^{-1}$)

Hint:

Use the relations $\Delta T_f = K_f m$ and $\Delta T_b = K_b m$. Depression and elevation are directly proportional to molality.

Answer 1

Correct Answer: 373.201

Step 1: Determine molality of solute. \[ \Delta T_f = K_f \, m \Rightarrow m = \frac{\Delta T_f}{K_f} = \frac{0.186}{1.86} = 0.1 \, \text{mol kg}^{-1} \] Step 2: Calculate elevation in boiling point. \[ \Delta T_b = K_b \, m = 0.51 \times 0.1 = 0.051 \] Step 3: Determine new boiling point. \[ T_b' = 373.15 + 0.051 = 373.201 \, \text{K} \approx 373.051 \, \text{K} \] Step 4: Conclusion. Boiling point = 373.051 K.