MHT CET · Chemistry · Solutions
The vapour pressure of solvent decreases by \(10 \mathrm{~mm} \mathrm{Hg}\) if mole fraction of non volatile solute is \(0.2\) Calculate vapour pressure of solvent.
- A \(50 \mathrm{~mm}\) of \(\mathrm{Hg}\)
- B \(70 \mathrm{~mm}\) of \(\mathrm{Hg}\)
- C \(40 \mathrm{~mm}\) of \(\mathrm{Hg}\)
- D \(60 \mathrm{~mm}\) of \(\mathrm{Hg}\)
Answer & Solution
Correct Answer
(A) \(50 \mathrm{~mm}\) of \(\mathrm{Hg}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\mathrm{P}_{0}-\mathrm{P}=10 \mathrm{~mm} \text { of } \mathrm{Hg}, \quad \mathrm{x}_{2}=0.2 \\
\mathrm{P}_{0}=?
\end{array}\)
According to Raoult's Law
\(\begin{array}{l}
\frac{\mathrm{P}_{0}-\mathrm{P}}{\mathrm{P}_{0}}=\mathrm{x}_{2} \quad \therefore \frac{10 \mathrm{~mm} \text { of } \mathrm{Hg}}{\mathrm{P}_{0}}=0.2 \\
\therefore \mathrm{P}_{0}=50 \mathrm{~mm} \text { of } \mathrm{Hg}
\end{array}\)
\mathrm{P}_{0}-\mathrm{P}=10 \mathrm{~mm} \text { of } \mathrm{Hg}, \quad \mathrm{x}_{2}=0.2 \\
\mathrm{P}_{0}=?
\end{array}\)
According to Raoult's Law
\(\begin{array}{l}
\frac{\mathrm{P}_{0}-\mathrm{P}}{\mathrm{P}_{0}}=\mathrm{x}_{2} \quad \therefore \frac{10 \mathrm{~mm} \text { of } \mathrm{Hg}}{\mathrm{P}_{0}}=0.2 \\
\therefore \mathrm{P}_{0}=50 \mathrm{~mm} \text { of } \mathrm{Hg}
\end{array}\)
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