CBSE Class 12 Chemistry 2023 Outside Delhi Set 1 Solved Paper

Question : 19 of 35

Marks: +1, -0

SECTION - B

The vapour pressure of pure liquid

X

and pure liquid

Y

25^{\circ}\mathrm{C}

are

120\,\text{mm}\,\text{Hg}

and

160\,\text{mm}\,\text{Hg}

respectively. If equal moles of

X

and

Y

are mixed to form an ideal solution, calculate the vapour pressure of the solution.

Solution: 👈: Video Solution

X=120\,\text{mm}\,\text{Hg}, P_A=P_A^0 \cdot x_A

Y=160\,\text{mm}\,\text{Hg}, P_B=P_B^0 \cdot x_B

\because

When equal amount of

X

and

Y

are mixed

x_A+x_B=1

Acc. to Raoutt's law

P_{\text{total}}=P_A+P_B

P_A^0 x_A + P_B^0 x_B

=(1-x_B)P_A^0+x_B P_B^0

P_{\text{Total}}=P_A^0+(P_B^0-P_A^0)x_B

=120+(160-120)\times 1

P_{\text{Total}}=120+40

=160\,\text{mm}\,\text{Hg}

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