Test Index

Liquid Solution

Section: Chemistry

© examsnet.com

Question : 2 of 29

Marks: +1, -0

At

T(K)

, the vapour pressure of

x

molal aqueous solution containing a non-volatile solute is 12.078 kPa . The vapour pressure of pure water at

T(K)

is 12.3 kPa . What is the value of

x

[AP EAPCET 24 May 2025 S1]

$\; 10$
$\; 1.018$
$\; 0.1018$
$\; 0.018$

Solution:

Mole fraction of solvent

\chi_{\;\text{solute }\;} \; = \; \frac{p_0-p_s}{p_0}

\Rightarrow \;\; \chi_{\;\text{Solute }\;} \; = \; \frac{12.3-12.078}{12.3} = 0.01805

Moles of water in 1000 g of water

= \; \frac{1000}{18} = 55.56 \text{ mol}

For dilute solution,

\chi_{\;\text{Solute }\;} = \frac{n_{\;\text{solute }\;}}{n_{\;\text{solute }\;}+n_{\;\text{water }\;}} \approx \frac{n_{\;\text{solute }\;}}{n_{\;\text{water }\;}}

Since molality

x

is moles of solute per 1 kg of solvent

\; x = n_{\;\text{solute }\;} \Rightarrow \;\; x_{\;\text{solute }\;} = \frac{x}{55.56}

\; x = 0.01805 \times 55.56 \approx 1.018

© examsnet.com

Go to Question:

Prev Question

Next Question