Consider the dissociation of the weak acid HX as given below
$$\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}$$
[$\mathrm{K}_{\mathrm{a}}$ : dissociation constant]
The osmotic pressure of $0.03 \mathrm{M}$ aqueous solution of $\mathrm{HX}$ at $300 \mathrm{~K}$ is _________ $\times 10^{-2}$ bar (nearest integer).
[Given : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$]
Answer (integer)
76
Solution
<p>$$\begin{aligned}
& \mathrm{Ka}=\frac{\mathrm{C} \alpha^2}{1-\alpha} \\
& 1.2 \times 10^{-5}=(0.03)\left(\alpha^2\right) \\
& \alpha=0.02 \\
& \pi=\mathrm{iCRT} \\
&=(1.02)(0.03)(0.083)(300) \\
&=0.76194 \\
&=76.194 \times 10^{-2} \mathrm{bar} \\
& \text { Nearest integer }=76
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
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