JEE Advanced · Chemistry · 16. Solutions
Vessel-1 contains \(\mathbf{w}_2 \mathrm{~g}\) of a non-volatile solute \(\mathbf{X}\) dissolved in \(\mathbf{w}_1 \mathrm{~g}\) of water. Vessel- 2 contains \(\mathbf{w}_2 \mathrm{~g}\) of another non-volatile solute \(\mathbf{Y}\) dissolved in \(\mathbf{w}_1 \mathrm{~g}\) of water. Both the vessels are at the same temperature and pressure. The molar mass of \(\mathbf{X}\) is \(80 \%\) of that of \(\mathbf{Y}\). The van't Hoff factor for \(\mathbf{X}\) is 1.2 times of that of \(\mathbf{Y}\) for their respective concentrations.
The elevation of boiling point for solution in Vessel-1 is ________. \(\%\) of the solution in Vessel-2.
- A 100
- B 150
- C 200
- D 250
Answer & Solution
Correct Answer
(B) 150
Step-by-step Solution
Detailed explanation
Vessel-I
\(\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{I}}=\mathrm{i}_{\mathrm{x}} \frac{\mathrm{w}_2}{\mathrm{M}_{\mathrm{x}}} \cdot \frac{1}{\mathrm{w}_1} \times 1000 \times \mathrm{K}_{\mathrm{b}}\)
\(\mathrm{M}_{\mathrm{X}}=\) Molar mass of ' X '
Vessel-II
\(\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{II}}=\mathrm{i}_{\mathrm{Y}} \frac{\mathrm{w}_2}{\mathrm{M}_{\mathrm{Y}}} \cdot \frac{1}{\mathrm{w}_1} \times 1000 \times \mathrm{K}_{\mathrm{b}}\)
\(M_Y=\) Molar mass of ' \(Y\) '
\(\begin{aligned}
& \frac{\left(\Delta T_b\right)_{\mathrm{I}}}{\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{II}}} \times 100=\frac{\mathrm{i}_{\mathrm{x}}}{\mathrm{i}_{\mathrm{Y}}} \cdot \frac{\mathrm{M}_{\mathrm{Y}}}{\mathrm{M}_{\mathrm{x}}} \times 100 \\
& =1.2 \times \frac{100}{80} \times 100 \\
& =150 \%
\end{aligned}\)
\(\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{I}}=\mathrm{i}_{\mathrm{x}} \frac{\mathrm{w}_2}{\mathrm{M}_{\mathrm{x}}} \cdot \frac{1}{\mathrm{w}_1} \times 1000 \times \mathrm{K}_{\mathrm{b}}\)
\(\mathrm{M}_{\mathrm{X}}=\) Molar mass of ' X '
Vessel-II
\(\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{II}}=\mathrm{i}_{\mathrm{Y}} \frac{\mathrm{w}_2}{\mathrm{M}_{\mathrm{Y}}} \cdot \frac{1}{\mathrm{w}_1} \times 1000 \times \mathrm{K}_{\mathrm{b}}\)
\(M_Y=\) Molar mass of ' \(Y\) '
\(\begin{aligned}
& \frac{\left(\Delta T_b\right)_{\mathrm{I}}}{\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{II}}} \times 100=\frac{\mathrm{i}_{\mathrm{x}}}{\mathrm{i}_{\mathrm{Y}}} \cdot \frac{\mathrm{M}_{\mathrm{Y}}}{\mathrm{M}_{\mathrm{x}}} \times 100 \\
& =1.2 \times \frac{100}{80} \times 100 \\
& =150 \%
\end{aligned}\)
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