MHT CET · Chemistry · Solutions
What is vapour pressure of a solution containing \(1 \mathrm{~mol}\) of a nonvolatile solute in \(36 \mathrm{~g}\) of water \(\left(\mathrm{P}_1^0=32 \mathrm{~mm} \mathrm{Hg}\right) ?\)
- A \(8.14 \mathrm{~mm} \mathrm{Hg}\)
- B \(12.31 \mathrm{~mm} \mathrm{Hg}\)
- C \(16.08 \mathrm{~mm} \mathrm{Hg}\)
- D \(21.44 \mathrm{~mm} \mathrm{Hg}\)
Answer & Solution
Correct Answer
(D) \(21.44 \mathrm{~mm} \mathrm{Hg}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{n}_2=1 \mathrm{~mol} \\
& \mathrm{n}_1=\frac{36}{18}=2 \mathrm{~mol}
\end{aligned}
\)
Relative lowering of vapour pressure
\(
=\frac{\mathrm{P}_1^0-\mathrm{P}_1}{\mathrm{P}_1^0}=x_2=\frac{\mathrm{n}_2}{\mathrm{n}_1+\mathrm{n}_2}
\)
\(\therefore \frac{32 \mathrm{~mm} \mathrm{Hg}-\mathrm{P}_1}{32 \mathrm{~mm} \mathrm{Hg}}=\frac{1}{3} \)
\( 96 \mathrm{~mm} \mathrm{Hg}-3 \mathrm{P}_1=32 \mathrm{~mm} \mathrm{Hg} \)
\( 64 \mathrm{~mm} \mathrm{Hg}=3 \mathrm{P}_1 \)
\( \therefore \mathrm{P}_1=21.33 \mathrm{~mm} \mathrm{Hg} \approx 21.44 \mathrm{~mm} \mathrm{Hg}\)
\begin{aligned}
& \mathrm{n}_2=1 \mathrm{~mol} \\
& \mathrm{n}_1=\frac{36}{18}=2 \mathrm{~mol}
\end{aligned}
\)
Relative lowering of vapour pressure
\(
=\frac{\mathrm{P}_1^0-\mathrm{P}_1}{\mathrm{P}_1^0}=x_2=\frac{\mathrm{n}_2}{\mathrm{n}_1+\mathrm{n}_2}
\)
\(\therefore \frac{32 \mathrm{~mm} \mathrm{Hg}-\mathrm{P}_1}{32 \mathrm{~mm} \mathrm{Hg}}=\frac{1}{3} \)
\( 96 \mathrm{~mm} \mathrm{Hg}-3 \mathrm{P}_1=32 \mathrm{~mm} \mathrm{Hg} \)
\( 64 \mathrm{~mm} \mathrm{Hg}=3 \mathrm{P}_1 \)
\( \therefore \mathrm{P}_1=21.33 \mathrm{~mm} \mathrm{Hg} \approx 21.44 \mathrm{~mm} \mathrm{Hg}\)
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