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QJEEM24Ch5f4f: JEE Main 2024 Chemistry - Class 11 JEEM24Ch5f4f Solution: Consider the dissociation of the weak acid HX as given below […

Consider the dissociation of the weak acid HX as given below

HX(aq)⇌H+(aq)+X−(aq),Ka=1.2×10−5\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}

[ Ka\mathrm{K}_{\mathrm{a}} : dissociation constant]

The osmotic pressure of 0.03M0.03 \mathrm{M} aqueous solution of HX\mathrm{HX} at 300 K300 \mathrm{~K} is _________ ×10−2\times 10^{-2} bar (nearest integer).

[Given : R=0.083 L bar mol−1 K−1\mathrm{R}=0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ]

Step-by-Step Verified Solution

amp;Ka=Cα21−αamp;1.2×10−5=(0.03)(α2)amp;α=0.02amp;π=iCRTamp;=(1.02)(0.03)(0.083)(300)amp;=0.76194amp;=76.194×10−2baramp; Nearest integer =76\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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