# A set of solutions is prepared using 180g of water as a solvent 10g of

Question:

A set of solutions is prepared using $180 \mathrm{~g}$ of water as a solvent and $10 \mathrm{~g}$ of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of $\mathrm{A}=100 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{B}=200 \mathrm{~g} \mathrm{~mol}^{-1}$; $\mathrm{C}=10,000 \mathrm{~g} \mathrm{~mol}^{-1}$ ]

1. $\mathrm{A}>\mathrm{B}>\mathrm{C}$

2. $A>C>B$

3. $\mathrm{C}>\mathrm{B}>\mathrm{A}$

4. $\mathrm{B}>\mathrm{C}>\mathrm{A}$

Correct Option: 1

Solution:

Relative lowering of V.P. $=\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}=\mathrm{x}_{\text {solute }}$

$\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{A}}=\frac{\frac{10}{100}}{\frac{10}{100}+\frac{180}{18}}:\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{B}}=\frac{\frac{10}{200}}{\frac{10}{200}+\frac{180}{18}}$

$\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{C}}=\frac{\frac{10}{10,000}}{\frac{10}{10,000}+\frac{180}{18}}:\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{A}}>\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{B}}>\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{C}}$