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Liquid Solution

Section: Chemistry

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Question : 11 of 29

Marks: +1, -0

At 300 K vapour pressure of a pure liquid, ' A ' is 70 mm Hg . It forms an ideal solution with another liquid ' B '. The mole fraction of B in the solution is 0.2 and total vapour pressure of solution is 84 mm Hg at same temperature. What is the vapour pressure (in mm) of pure liquid B at 300 K ?

[AP EAPCET 21 May 2025 S2]

$140$
$70$
$280$
$560$

Solution:

Given data

P_A^0=70\,\text{mm Hg}

Ideal solution

\rightarrow

obeys Raoult's law

x_B=0.2 \Rightarrow x_A=1-0.2=0.8

Total vapour pressure of solution

P_{\;\text{total }\;}=84\,\text{mm Hg}

\;\text{ Temperature = }\;300\,\text{K}\;\text{ (constant) }\;

We are asked to find

P_B^0

(vapour pressure of pure liquid

B

).

Step 1: Apply Raoult's Law

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

Substitute known values:

84=(0.8)(70)+(0.2)(P_B^0)

Step 2: Simplify

\;84=56+0.2 P_B^0

\;84-56=0.2 P_B^0

\;28=0.2 P_B^0

Step 3: Solve for

P_B^0

P_B^0=\;\frac{28}{0.2}=140\,\text{mm Hg}

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