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Section: Chemistry

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Question : 3 of 23

Marks: +1, -0

50\ \mathrm{mL}

of 0.2 molal urea solution (density

=1.012\ \mathrm{g}\ \mathrm{mL}^{-1}

300\ \mathrm{K}

) is mixed with

250\ \mathrm{mL}

of a solution containing

0.06\ \mathrm{g}

of urea. Both the solutions were prepared in the same solvent. The osmotic pressure (in torr) of the resulting at

300\ \mathrm{K}

[Use : Molar mass of urea

=60\ \mathrm{g}\ \mathrm{mol}^{-1}

; gas constant,

R=62\ \mathrm{L}_{\text{torr K}}^{-1}\ \mathrm{mol}^{-1}

\Delta_{\text{mix }} H=0,\ \Delta_{\text{mix }} V=0

[JEE Adv 2023 P2]

Your Answer:

Solution:

Mole of urea

=0.2

Weight of urea

=0.2\times 60=12\ \mathrm{g}

Weight of solvent

=1000\ \mathrm{g}

Weight of solution

=1012\ \mathrm{g}

\therefore

Volume of solution

=\frac{1012}{1.012}=1000\ \mathrm{ml}

\because 1000\ \mathrm{ml}

solution contain 0.2 mole

\therefore 50\ \mathrm{ml}

solution contain

=\frac{0.2\times 50}{1000}=0.01

Mole of urea in other solution

=\frac{0.06}{60}=0.001

\therefore\ \text{ Concentration of solution }\ =\frac{0.01+0.001}{\frac{300}{1000}}

=0.0366

\therefore\ \pi = CRT

=0.0366\times 62\times 300

=682

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