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 M aqueous solution of HX at 300 K is \_\_\_\_ $\times 10^{-2}$ bar (nearest integer). [Given : $\mathrm{R}=0.083 \mathrm{~L}$ bar $\mathrm{Mol}^{-1} \mathrm{~K}^{-1}$ ]

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Solutions

Consider the dissociation of the weak acid HX as given below

HX (aq) ⇌ H^{+} (aq) + X^{-} (aq), Ka = 1.2 \times 1 0^{- 5}

[

K_{a}

: dissociation constant]
The osmotic pressure of 0.03 M aqueous solution of HX at 300 K is ____

\times 1 0^{- 2}

bar (nearest integer).
[Given :

R = 0.083 L

bar

Mol^{- 1} K^{- 1}

]

Answer: 76

Solution

For the weak acid HX:

[H^{+}] = [X^{-}] = x = K_{a} C = 1.2 \times 1 0^{- 5} \times 0.03 = 6 \times 1 0^{- 4}

M.
Total solute concentration

= C + x = 0.03 + 6 \times 1 0^{- 4} = 0.0306

π = C_{t o t a l} R T = 0.0306 \times 0.083 \times 300 = 0.762 bar = 76.2 \times 1 0^{- 2} bar \approx 76 \times 1 0^{- 2} bar .

The value is

76.

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Integer answer

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