1. Tardigrade
  2. Question
  3. Chemistry
  4. A set of solutions is prepared using 180 g of water as a solvent and 10 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 .A=100 g mol -1 ; B =200 g mol -1 ; C =10,000 g mol -1]

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A set of solutions is prepared using $180 \,g$ of water as a solvent and $10 \,g$ 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 $\left.A=100 g mol ^{-1} ; B =200 g mol ^{-1} ; C =10,000 g mol ^{-1}\right]$

JEE MainJEE Main 2020Solutions

Solution:

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

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

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