Chemistry · 2024
JEE Main · 6 April 2024, Shift 1 · Q84
Consider the dissociation of the weak acid HX as given below HX ( aq ) ⇌ H^+( aq )+ X^-( aq ), Ka =1.2 × 10^-5 [ K_a:. dissociation constant ] The…
Consider the dissociation of the weak acid HX as given below
$$\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}
$$
$\displaystyle \left[\mathrm{K}_{\mathrm{a}}:\right.$ dissociation constant $\displaystyle ]$
The osmotic pressure of $\displaystyle 0.03$ M aqueous solution of HX at $\displaystyle 300$ K is $\displaystyle \_\_\_\_$
$\displaystyle \times 10^{-2}$ bar (nearest integer).
[Given : $\displaystyle \mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$ ]
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From NTA’s final answer key for this paper.
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JEE Main 2024 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.