MHT CET · Chemistry · Solutions
Calculate vapour pressure of a solution containing mixture of 2 moles of volatile liquid A and 3 moles of volatile liquid B at room temperature.
\(\left(\mathrm{P}_{\mathrm{A}}^{\circ}=420, \mathrm{P}_{\mathrm{B}}^{\circ}=610 \mathrm{~mm} \mathrm{Hg}\right)\)
- A 600 mm Hg
- B 570 mm Hg
- C 534 mm Hg
- D 480 mm Hg
Answer & Solution
Correct Answer
(C) 534 mm Hg
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{x}_1=\frac{\mathrm{n}_1}{\mathrm{n}_1+\mathrm{n}_2}=\frac{2}{2+3}=0.4 \\
& \mathrm{x}_2=\frac{\mathrm{n}_1}{\mathrm{n}_1+\mathrm{n}_2}=\frac{3}{2+3}=0.6 \\
& \mathrm{P}_{\text {solution }}=\mathrm{P}_{\mathrm{A}}^0 \mathrm{x}_1+\mathrm{P}_{\mathrm{B}}^0 \mathrm{x}_2 \\
& =420 \times 0.4+610 \times 0.6 \\
& =168+366 \\
& =534 \mathrm{~mm} \mathrm{Hg}
\end{aligned}\)
& \mathrm{x}_1=\frac{\mathrm{n}_1}{\mathrm{n}_1+\mathrm{n}_2}=\frac{2}{2+3}=0.4 \\
& \mathrm{x}_2=\frac{\mathrm{n}_1}{\mathrm{n}_1+\mathrm{n}_2}=\frac{3}{2+3}=0.6 \\
& \mathrm{P}_{\text {solution }}=\mathrm{P}_{\mathrm{A}}^0 \mathrm{x}_1+\mathrm{P}_{\mathrm{B}}^0 \mathrm{x}_2 \\
& =420 \times 0.4+610 \times 0.6 \\
& =168+366 \\
& =534 \mathrm{~mm} \mathrm{Hg}
\end{aligned}\)
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