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 $\mathrm{A}=100 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{B}=200 \mathrm{~g} \mathrm{~mol}^{-1}$; $\mathrm{C}=10,000 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
Correct Option: 1
Relative lowering of V.P. $=\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}=\mathrm{x}_{\text {solute }}$
$\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{A}}=\frac{\frac{10}{100}}{\frac{10}{100}+\frac{180}{18}}:\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{B}}=\frac{\frac{10}{200}}{\frac{10}{200}+\frac{180}{18}}$
$\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{C}}=\frac{\frac{10}{10,000}}{\frac{10}{10,000}+\frac{180}{18}}:\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{A}}>\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{B}}>\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{C}}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.