JEE MAIN - Chemistry (2024 - 6th April Morning Shift - No. 23)
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
76
Explanation
$$\begin{aligned}
& \mathrm{Ka}=\frac{\mathrm{C} \alpha^2}{1-\alpha} \\
& 1.2 \times 10^{-5}=(0.03)\left(\alpha^2\right) \\
& \alpha=0.02 \\
& \pi=\mathrm{iCRT} \\
&=(1.02)(0.03)(0.083)(300) \\
&=0.76194 \\
&=76.194 \times 10^{-2} \mathrm{bar} \\
& \text { Nearest integer }=76
\end{aligned}$$
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