JEE Main 2021 · 24 Feb · Shift 1 · Chemistry Q55

← JEE Main 2021 · 24 Feb · Shift 1

ChemistryInteger+4 / -1Question 55 of 90

When 9.45 g of ClCH $_2$ COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of ClCH $_2$ COOH is $x\times10^{-3}$ . The value of $x$ is _______. (Rounded off to the nearest integer)

[ $K_{f(\mathrm{H_2O})}$ = 1.86 K kg mol $^{-1}$ ]

Your answer

Answer and worked solution

Official answer: 36

van 't Hoff factor. Molality $\frac{0.1}{0.5}=0.2$ ; $0.5=1.86\times0.2\,i$ gives $i=1.344$ , $\alpha=0.344$ .

Ka $\approx\dfrac{0.2\alpha^2}{1-\alpha}=\dfrac{0.2\times0.118}{0.656}\approx0.036=36\times10^{-3}$ .

Answer: 36

Topics:van't Hoff factor pH and ionisation

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JEE Main 2021 · 24 Feb · Shift 1, Chemistry question 55