MHT CET · Chemistry · Solutions
What is vapour pressure of a solution containing \(0.1 \mathrm{~mol}\) of non-volatile solute dissolved in \(16.2 \mathrm{~g}\) of water?
\(
\left(\mathrm{P}_1^{\circ}=32 \mathrm{~mm} \mathrm{Hg}\right)
\)
- A \(21.6 \mathrm{~mm} \mathrm{Hg}\)
- B \(28.8 \mathrm{~mm} \mathrm{Hg}\)
- C \(15.7 \mathrm{~mm} \mathrm{Hg}\)
- D \(18.1 \mathrm{~mm} \mathrm{Hg}\)
Answer & Solution
Correct Answer
(B) \(28.8 \mathrm{~mm} \mathrm{Hg}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \frac{\mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}^{\circ}}=\mathrm{X}_{\text {solute }} \\
& \frac{32-\mathrm{P}_{\mathrm{s}}}{32}=\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solute }}+\mathrm{n}_{\text {solvent }}} \\
& \mathrm{n}_{\text {solvent }}=\mathrm{n}_{\text {water }}=\frac{16.2}{18}=0.9 \\
& \frac{32-\mathrm{P}_{\mathrm{s}}}{32}=\frac{0.1}{0.1+0.9}=0.1 \\
& 32-\mathrm{P}_{\mathrm{s}}=32 \times 0.1 \\
& \mathrm{P}_{\mathrm{s}}=32-3.2 \\
& =28.8 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
\)
\begin{aligned}
& \frac{\mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}^{\circ}}=\mathrm{X}_{\text {solute }} \\
& \frac{32-\mathrm{P}_{\mathrm{s}}}{32}=\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solute }}+\mathrm{n}_{\text {solvent }}} \\
& \mathrm{n}_{\text {solvent }}=\mathrm{n}_{\text {water }}=\frac{16.2}{18}=0.9 \\
& \frac{32-\mathrm{P}_{\mathrm{s}}}{32}=\frac{0.1}{0.1+0.9}=0.1 \\
& 32-\mathrm{P}_{\mathrm{s}}=32 \times 0.1 \\
& \mathrm{P}_{\mathrm{s}}=32-3.2 \\
& =28.8 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
\)
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