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uestion

Asked in: JEE Main-2019

At room temperature, a dilute solution of urea is prepared by dissolving 0.60g of urea

in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mmHg,

lowering of vapour pressure will be:

(molar mass of urea = 60g/mol)

A.

0.027 mmHg

B.

0.028 mmHg

C.

0.017 mmHg

D.

0.031 mmHg

uestion

Asked in: JEE Main-2019

At room temperature, a dilute solution of urea is prepared by dissolving 0.60g of urea

in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mmHg,

lowering of vapour pressure will be:

(molar mass of urea = 60g/mol)

A.

0.027 mmHg

B.

0.028 mmHg

C.

0.017 mmHg

D.

0.031 mmHgv

uestion

Asked in: JEE Main-2019

At room temperature, a dilute solution of urea is prepared by dissolving 0.60g of urea

in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mmHg,

lowering of vapour pressure will be:

(molar mass of urea = 60g/mol)

A.

0.027 mmHg

B.

0.028 mmHg

C.

0.017 mmHg

D.

0.031 mmHg

Answers (1)

best_answer

Lowering Of vapor Pressure:

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

Now,

\begin{array}{l}{\therefore \Delta p=35 \times \frac{0.6 / 60}{\frac{0.6}{60}+\frac{360}{18}}} \\ {=35 \times \frac{0.1}{01+20}=35 \times \frac{.01}{20.01}} \\ {=.017 \mathrm{mm} \mathrm{Hg}}\end{array}

Posted by

Pankaj Sanodiya

View full answer

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