At room temperature, a dilute solution of urea is prepared by dissolving 0.60 of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be. (molar mass of urea = 60 g mol–1)

Solutions PYQ — JEE Main 2019 · 10 Apr · Shift 1 · Q9 with Solution

View written solutionFree

Correct answer: B

For a non-volatile solute,

\Delta p = p^0 x_{\text{solute}}

where:

For a dilute solution, this is the standard relation.

Given mass of urea $= 0.60\,\text{g}$ and molar mass $= 60\,\text{g mol}^{-1}$ ,

n_{\text{urea}} = \frac{0.60}{60} = 0.01\,\text{mol}

Given mass of water $= 360\,\text{g}$ and molar mass of water $= 18\,\text{g mol}^{-1}$ ,

n_{\text{water}} = \frac{360}{18} = 20\,\text{mol}

x_{\text{urea}} = \frac{n_{\text{urea}}}{n_{\text{urea}} + n_{\text{water}}} = \frac{0.01}{20.01}

Since the solution is dilute,

x_{\text{urea}} \approx \frac{0.01}{20} = 5 \times 10^{-4}

More accurately,

x_{\text{urea}} = 4.9975 \times 10^{-4}

\Delta p = p^0 x_{\text{solute}} = 35 \times 4.9975 \times 10^{-4}

\Delta p \approx 0.01749\,\text{mmHg}

So,

\Delta p \approx 0.017\,\text{mmHg}

The correct option is:

\boxed{\text{B: } 0.017\,\text{mmHg}}