Question

If the depression in freezing point of an aqueous solution containing a solute, which is neither dissociated nor associated, is $aK$ with $K_f = b \ \text{K} \cdot \text{kg mol}^{-1}$, what would be the elevation in boiling point ($X$) for this solution if its $K_b = K \ \text{K} \cdot \text{kg mol}^{-1}$?

• A. X = 2c $\times$ $\frac{b}{a}$
• B. X = c $\times$ $\frac{a}{2b}$
• C. X = c $\times$ $\frac{a}{b}$
• D. X = c $\times$ $\frac{b}{a}$

Hint:

When solving questions on colligative properties like freezing point depression and boiling point elevation, remember that the formula involves the cryoscopic or ebullioscopic constant and the molality of the solution. The relationships between them are proportional.

Answer 1

The Correct Option is C

We can relate the depression in freezing point and the elevation in boiling point by the following formulas: \[ \Delta T_f = K_f \times m \quad \text{and} \quad \Delta T_b = K_b \times m \] where $\Delta T_f$ and $\Delta T_b$ are the changes in freezing and boiling points, respectively, $K_f$ and $K_b$ are the cryoscopic and ebullioscopic constants, and $m$ is the molality of the solution. We are given that the depression in freezing point is $aK$, so: \[ \Delta T_f = aK \] From the formula for depression in freezing point, we have: \[ aK = K_f \times m \] which gives the molality of the solution: \[ m = \frac{aK}{K_f} \] Now, for the elevation in boiling point, we have the equation: \[ \Delta T_b = K_b \times m \] Substituting for $m$, we get: \[ \Delta T_b = K_b \times \frac{aK}{K_f} \] Thus, the elevation in boiling point ($X$) is: \[ X = c \times \frac{a}{b} \] Therefore, the correct relationship is $X = c \times \frac{a}{b}$, which corresponds to option (C)