Solutions and Colligative Properties 2 Question 2
2. $1 \mathrm{~g}$ of a non-volatile, non-electrolyte solute is dissolved in $100 \mathrm{~g}$ of two different solvents $A$ and $B$, whose ebullisocopic constants are in the ratio of $1: 5$. The ratio of the elevation in their boiling points, $\frac{\Delta T_{b}(A)}{\Delta T_{b}(B)}$, is
(2019 Main, 10 April II)
(a) $5: 1$
(b) $10: 1$
(c) $1: 5$
(d) $1: 0.2$
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Solution:
- The expression of elevation of boiling point,
$$ \begin{aligned} \Delta T_{b} & =K_{b} \times m \times i \ & =k_{b} \times \frac{w_{2} \times 1000}{M_{2} \times w_{1}} \times i \end{aligned} $$
where, $m=$ molality
$i=$ van’t Hoff factor $=1$ (for non-electrolyte/non-associable)
$w_{2}=$ mass of solute in $\mathrm{g}=1 \mathrm{~g}$ (present in both of the solutions)
$M_{2}=$ molar mass of solute in $\mathrm{g} \mathrm{mol}^{-1}$ (same solute in both of the solutions)
$w_{1}=$ mass of solvent in $\mathrm{g}=100 \mathrm{~g}$ (for both of the solvents $A$ and $B$ )
$K_{b}=$ ebullioscopic constant
So, the expression becomes,
$$ \begin{aligned} & \Delta T_{b} \propto K_{b} \ & \Rightarrow \quad \frac{\Delta T_{b}(A)}{\Delta T_{b}(B)}=\frac{K_{b}(A)}{K_{b}(B)}=\frac{1}{5} \quad\left[\text { Given } \frac{K_{b}(A)}{K_{b}(B)}=\frac{1}{5}\right] \end{aligned} $$
