MHT CET · Chemistry · Solutions
What is vapour pressure of a solution when \(2 \mathrm{~mol}\) of a non-volatile solute are dissolved in \(20 \mathrm{~mol}\) of water? \(\left(\mathrm{P}_1^{\circ}=32 \mathrm{~mm} \mathrm{Hg}\right)\)
- A \(29.1 \mathrm{~mm} \mathrm{Hg}\)
- B \(12 \mathrm{~mm} \mathrm{Hg}\)
- C \(6 \mathrm{~mm} \mathrm{Hg}\)
- D \(9 \mathrm{~mm} \mathrm{Hg}\)
Answer & Solution
Correct Answer
(A) \(29.1 \mathrm{~mm} \mathrm{Hg}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{\mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}^{\circ}}=\mathrm{X}_{\text {solute }}=\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solute }}+\mathrm{n}_{\text {solvent }}} \\ & \frac{32-\mathrm{P}_{\mathrm{s}}}{32}=\frac{2}{2+20} \\ & 32-\mathrm{P}_{\mathrm{s}}=\frac{2}{22} \times 32=2.90 \\ & \mathrm{P}_{\mathrm{s}}=32-2.90 \\ & =29.1 \mathrm{~mm} \mathrm{Hg}\end{aligned}\)
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