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JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties

Consider the dissociation of the weak acid \(\mathrm{HX}\) as given below \(\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}\) \(\left[\mathrm{K}_{\mathrm{n}}:\right.\) 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). \(\left[\right.\) Given : \(\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{Mol}^{-1} \mathrm{~K}^{-1}\) ]

  1. A \(76\)
  2. B \(77\)
  3. C \(79\)
  4. D \(80\)
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Correct Answer

(A) \(76\)

Step-by-step Solution

Detailed explanation

\(\mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-} \quad \mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5}\) \(0.03 \mathrm{M}\) \(0.03-\mathrm{x} \quad \mathrm{x} \quad \mathrm{x}\) \(\mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5}=\frac{\mathrm{x}^2}{0.03-\mathrm{x}}\)…
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