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Liquid Solution

Section: Chemistry

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Question : 7 of 29

Marks: +1, -0

At

298\ \text{K}, 0.714

moles of liquid A is dissolved in 5.555 moles of liquid B . The vapour pressure of the resultant solution is 475 torr. The vapour pressure of pure liquid

A

at the same temperature is 280.7 torr. What is the vapour pressure of pure liquid B in torr ?

[AP EAPCET 22 May 2025 S2]

$486$
$550$
514
500

Solution:

Moles of

A=0.714\ \text{mol}

Moles of

B=5.555\ \text{mol}

P_{\;\text{solution}\;}=475

torr

P_A^0=280.7

torr

P_B^0=

?

Step 1. Mole fractions

\;x_A=\;\frac{0.714}{0.714+5.555}=\;\frac{0.714}{6.269}=0.1138

\;x_B=1-x_A=0.8862

Step 2. Raoult's law

For an ideal solution:

P_{\;\text{total}\;}=P_A+P_B=x_A P_A^0+x_B P_B^0

We know everything except

P_B^0

.

So:

475=(0.1138)(280.7)+(0.8862) P_B^0

Step 3. Solve for

P_B^0

Compute first term:

0.1138 \times 280.7=31.93

Thus

\;475=31.93+0.8862 P_B^0

\;0.8862 P_B^0=475-31.93=443.07

\;P_B^0=\;\frac{443.07}{0.8862}=500.1

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