Solutions and Colligative Properties - Result Question 2-1
2. $1$ g of a non-volatile, non-electrolyte solute is dissolved in $100$ 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, $\dfrac{\Delta T_b(A)}{\Delta T_b(B)}$, is
(2019 Main, 10 April II)
(a) $5: 1$
(b) $10: 1$
(c) $1: 5$
(d) $1: 0.2$
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Answer:
Correct Answer: 2. (c)
Solution:
The expression of elevation of boiling point,
$\Delta T_b =K_b \times m \times i $
$ =k_b \times \dfrac{w_2 \times 1000}{M_2 \times w_1} \times i$
$ \text { where, } m=\text { molality } $
$ i=\text { van’t Hoff factor }=1 \text { (for non-electrolyte/non-associable) } $
$ w_2=\text { mass of solute in } \mathrm{g}=1 \mathrm{~g} \text { (present in both of the solutions) } $
$ M_2=\text { molar mass of solute in } \mathrm{g} $ $\mathrm{mol}^{-1} \text { (same solute in both of the solutions) } $
$ w_1=\text { mass of solvent in } \mathrm{g}=100 \mathrm{~g} \text { (for both of the solvents } A \text { and } B \text { ) } $
$ K_b=\text { ebullioscopic constant } $
$ \text { So, the expression becomes, } $
$ \Delta T_b \propto K_b $
$ \Rightarrow \dfrac{\Delta T_b(A)}{\Delta T_b(B)}=\dfrac{K_b(A)}{K_b(B)}=\dfrac{1}{5} \quad\left[\text { Given } \dfrac{K_b(A)}{K_b(B)}=\dfrac{1}{5}\right] $
BETA
