The boiling point and freezing point of a solvent 'A' are and , respectively. and values of the solv — Solutions and Colligative Properties Chemistry Question
Question
The boiling point and freezing point of a solvent 'A' are $90.0^\circ\text{C}$ and $3.5^\circ\text{C}$, respectively. $K_f$ and $K_b$ values of the solvent are 17.5 and 5.0 K-kg/mol, respectively. What is the boiling point of a solution of 'B' (non-volatile, non-electrolyte solute) in 'A', if the solution freezes at $2.8^\circ\text{C}$?
Answer: C
💡 Solution & Explanation
The freezing point depression $\Delta T_f = 3.5 - 2.8 = 0.7^\circ\text{C}$. Molality $m = \Delta T_f / K_f = 0.7 / 17.5 = 0.04 \text{ m}$. The boiling point elevation $\Delta T_b = K_b \times m = 5.0 \times 0.04 = 0.2^\circ\text{C}$. The new boiling point is $90.0 + 0.2 = 90.2^\circ\text{C}$. Therefore, correct answer is C.
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