Question

If $P_{0}$ and $P_{S}$ are the vapour pressure of solvent and solution, respectively and $N_{1}$ and $N_{2}$ are the moles of solute and solvent then

• A. $\frac{P_{0}-P_{S}}{P_{0}}=\frac{N_{1}}{N_{1}+N_{2}}$
• B. $\frac{P_{0}-P_{S}}{P_{S}}=\frac{N_{1}}{N_{2}}$
• C. $P_{S}=P_{0} \cdot \frac{N_{2}}{N_{1}+N_{2}}$
• D. All of these.

Answer 1

Answer: (D)

According to Raoult's law,

$\frac{P_{0}-P_{S}}{P_{0}}=\frac{N_{1}}{N_{1}+N_{2}} ; $

$ \therefore 1-\frac{P_{S}}{P_{0}}=\frac{N_{1}}{N_{1}+N_{2}}$

or $\frac{P_{S}}{P_{0}}=1-\frac{N_{1}}{N_{1}+N_{2}}=\frac{N_{2}}{N_{1}+N_{2}}$ or $P_{S}=P_{0} \times \frac{N_{2}}{N_{1}+N_{2}}$

Again, $\frac{P_{0}-P_{S}}{P_{0}}=\frac{N_{1}}{N_{1}+N_{2}}$

$\therefore \frac{P_{0}}{P_{0}-P_{S}}=\frac{N_{1}+N_{2}}{N_{1}}=1+\frac{N_{2}}{N_{1}}$

or $\frac{P_{S}}{P_{0}-P_{S}}=\frac{N_{2}}{N_{1}} $

or $ \frac{P_{0}-P_{S}}{P_{S}}=\frac{N_{1}}{N_{2}}$