A set of solutions is prepared using 180 g of water as a solvent and 10 g of different nonvolatile 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]

Solutions PYQ — JEE Main 2020 · 6 Sep · Shift 2 · Q15 with Solution

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Correct answer: C

Use Raoult’s law for relative lowering of vapour pressure

For a nonvolatile solute,

\frac{p^0 - p}{p^0} = x_{\text{solute}}

where $x_{\text{solute}}$ is the mole fraction of solute.

Since all solutions are made with the same mass of solvent $(180\,\text{g water})$ and same mass of solute $(10\,\text{g})$ , the relative lowering depends on the number of moles of solute.

Moles of solvent

Molar mass of water $= 18\,\text{g mol}^{-1}$

n_{\text{water}} = \frac{180}{18} = 10\,\text{mol}

Moles of each solute

For $A$ :

n_A = \frac{10}{100} = 0.1\,\text{mol}

For $B$ :

n_B = \frac{10}{200} = 0.05\,\text{mol}

For $C$ :

n_C = \frac{10}{10000} = 0.001\,\text{mol}

Compare mole fractions of solutes

x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{water}}}

For $A$ :

x_A = \frac{0.1}{10.1}

For $B$ :

x_B = \frac{0.05}{10.05}

For $C$ :

x_C = \frac{0.001}{10.001}

Clearly,

x_A > x_B > x_C

Hence, the relative lowering of vapour pressure is in the order:

A > B > C

Match with options

This corresponds to Option C.

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