MHT CET · Chemistry · Solutions
What is the vapour pressure of a solution containing 0.1 mol of non volatile solute dissolved in 16.2 g water? \(\left(\mathrm{P}_1^0=24 \mathrm{mmHg}\right.\), molar mass of water \(18 \mathrm{~g} \mathrm{~mol}^{-1}\) )
- A 12.4 mmHg
- B 18.1 mm Hg
- C 15.7 mmHg
- D \(21.3 \mathrm{mmHg}\)
Answer & Solution
Correct Answer
(D) \(21.3 \mathrm{mmHg}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \frac{\mathrm{P}_1^0-\mathrm{P}_1}{\mathrm{P}_1^0}=\frac{\mathrm{n}_2 \mathrm{M}_1}{\mathrm{~W}_1} \\ \therefore \quad & \frac{24-\mathrm{P}_1}{24}=\frac{0.1 \times 18}{16.2} \\ \therefore \quad & \mathrm{P}_1=21.33 \mathrm{~mm} \mathrm{Hg}\end{array}\)
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