Q12 / 180 · COMEDK 2025 (10 May Shift 1) · Chemistry
For a 1.0 molal solution containing the non-volatile solute Urea, the elevation in boiling point is 2.0 K while the depression in freezing point in a 3.0 molal solution having the same solvent is $4.0 K$. If the ratio $\dfrac{K_b}{K_f}=\dfrac{1}{X}$, what is the value of $X$ ?
- Correct answerA$\dfrac{2}{3}$
- B$\dfrac{3}{2}$
- C$\dfrac{1}{2}$
- D$\dfrac{1}{4}$
Solution
The elevation in boiling point is given by $\Delta T_b = K_b \times m_1$, where $m_1 = 1.0$ molal and $\Delta T_b = 2.0$ K. Thus, $2.0 = K_b \times 1.0$, which implies $K_b = 2.0$ K kg mol$^{-1}$.
The depression in freezing point is given by $\Delta T_f = K_f \times m_2$, where $m_2 = 3.0$ molal and $\Delta T_f = 4.0$ K. Thus, $4.0 = K_f \times 3.0$, which implies $K_f = \dfrac{4.0}{3.0} = \dfrac{4}{3}$ K kg mol$^{-1}$.
The ratio $\dfrac{K_b}{K_f}$ is calculated as $\dfrac{2.0}{4/3} = 2.0 \times \dfrac{3}{4} = \dfrac{6}{4} = \dfrac{3}{2}$.
Given $\dfrac{K_b}{K_f} = \dfrac{1}{X}$, we have $\dfrac{3}{2} = \dfrac{1}{X}$, which implies $X = \dfrac{2}{3}$.
Answer: $\dfrac{2}{3}$
The depression in freezing point is given by $\Delta T_f = K_f \times m_2$, where $m_2 = 3.0$ molal and $\Delta T_f = 4.0$ K. Thus, $4.0 = K_f \times 3.0$, which implies $K_f = \dfrac{4.0}{3.0} = \dfrac{4}{3}$ K kg mol$^{-1}$.
The ratio $\dfrac{K_b}{K_f}$ is calculated as $\dfrac{2.0}{4/3} = 2.0 \times \dfrac{3}{4} = \dfrac{6}{4} = \dfrac{3}{2}$.
Given $\dfrac{K_b}{K_f} = \dfrac{1}{X}$, we have $\dfrac{3}{2} = \dfrac{1}{X}$, which implies $X = \dfrac{2}{3}$.
Answer: $\dfrac{2}{3}$
Hence, the correct answer is option: 'A'.