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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2024 · 6 April · Morning Shift · Chemistry · Solutions

Consider the dissociation of the weak acid ( HX as given below (...

Consider the dissociation of the weak acid (HX\mathrm{HX} as given below (HX(aq)⇌H+(aq)+X−(aq),Ka=\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}), \mathrm{Ka}=1.2 ×\times10^{-5}) ([Ka\left[\mathrm{K}_{\mathrm{a}}\right. : dissociation constant ]] The osmotic pressure of (0.03 M\mathrm{M} aqueous solution of (HX\mathrm{HX} at (300  K\mathrm{~K} is _______ (×\times10^{-2}) bar (nearest integer). [Given : (R=\mathrm{R}=0.083  Lbarmol−1 K−1\mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1}]

Correct answer

0

Step-by-step solution

HX⇌H++X−Ka=1.2×10−50.03M0.03−xxxKa=1.2×10−5=x20.03−x0.03−x≈0.03(Ka is very small )x20.03=1.2×10−5x=6×10−4\begin{aligned} & \mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-} \quad \mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5} \\ & 0.03 \mathrm{M} \\ & 0.03-\mathrm{x} \quad \mathrm{x} \quad \mathrm{x} \\ & \mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5}=\frac{\mathrm{x}^2}{0.03-\mathrm{x}} \\ & 0.03-\mathrm{x} \approx 0.03\left(\mathrm{ K}_{\mathrm{a}} \text{ is very small }\right) \\ & \frac{\mathrm{x}^2}{0.03}=1.2 \times 10^{-5} \\ & \mathrm{x}=6 \times 10^{-4} \end{aligned}Final solution : 0.03−x+x+x0.03-x+x+x=0.03+x=0.03+6×10−4Π=(0.03+(6×10−4))×0.083×300=76.19×10−2≈76×10−2\begin{aligned} & =0.03+x=0.03+6 \times 10^{-4} \\ & \Pi=\left(0.03+\left(6 \times 10^{-4}\right)\right) \times 0.083 \times 300 \\ & =76.19 \times 10^{-2} \approx 76 \times 10^{-2} \end{aligned}$$
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