(B) $\Delta T_{f} = i \times K_{f} \times m$ $\therefore i = \frac{\Delta T_{f}}{K_{f} \times m} = \frac{0.19}{1.86 \times 0.1} = 1.0215 \approx 1.02$

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(B) $\Delta T_{f} = i \times K_{f} \times m$ $\therefore i = \frac{\Delta T_{f}}{K_{f} \times m} = \frac{0.19}{1.86 \times 0.1} = 1.0215 \approx 1.02$ Again from, $\alpha = \frac{i-1}{n-1} = \frac{1.02-1}{2-1} = 0.02$ For $CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+}$ $K_{a} = C \alpha^{2}$ $= 0.1 \times (0.02)^{2} = 0.1 \times 4 \times 10^{-4} = 4 \times 10^{-5}$

Class 12 Chemistry Solutions Abnormal molecular mass

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The measured freezing point depression for a $0.1 \ m$ aqueous $CH_{3}COOH$ solution is $0.19^{\circ} C$. The acid dissociation constant $K_{a}$ at this concentration will be (Given, $K_{f}$ the molal cryoscopic constant $= 1.86 \ K \ kg \ mol^{-1}$)

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A
$4.76 \times 10^{-5}$
B
$4 \times 10^{-5}$
C
$8 \times 10^{-5}$
D
$2 \times 10^{-5}$

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Three particles of a solute $A$ associate in benzene to form a trimer $A_3$. Calculate the freezing point of a $0.25 \ m$ solution if the degree of association of solute $A$ is $0.80$. The freezing point of benzene is $5.5 \ ^oC$ and $K_f$ is $5.12 \ K \ kg \ mol^{-1}$. (Answer in $K$)

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For a $1.5 \, m$ aqueous solution of $Ca(NO_3)_2$,the observed molar mass is $65.4 \, g \, mol^{-1}$ and the theoretical molar mass is $164 \, g \, mol^{-1}$. The degree of dissociation is .............

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In a solvent,$50\,\%$ of an acid $HA$ dimerizes and the rest dissociates. The van't Hoff factor of the acid is $.....\times 10^{-2}$. (Round off to the nearest integer)

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$A$ $0.5\%$ solution of potassium chloride was found to freeze at $-0.24^{\circ} C$. The percentage dissociation of potassium chloride is .... (Nearest integer)
(Molal depression constant for water is $1.80\, K\, kg\, mol^{-1}$ and molar mass of $KCl$ is $74.6\, g\, mol^{-1}$)

DifficultJEE MAIN 2022

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$A$ solute $A$ undergoes association in a medium as $nA \rightleftharpoons A_n$. If the degree of association is $\alpha$,then the van't Hoff factor $i$ is given by:

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The measured freezing point depression for a $0.1 \ m$ aqueous $CH_{3}COOH$ solution is $0.19^{\circ} C$. The acid dissociation constant $K_{a}$ at this concentration will be (Given, $K_{f}$ the molal cryoscopic constant $= 1.86 \ K \ kg \ mol^{-1}$)

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Three particles of a solute $A$ associate in benzene to form a trimer $A_3$. Calculate the freezing point of a $0.25 \ m$ solution if the degree of association of solute $A$ is $0.80$. The freezing point of benzene is $5.5 \ ^oC$ and $K_f$ is $5.12 \ K \ kg \ mol^{-1}$. (Answer in $K$)

For a $1.5 \, m$ aqueous solution of $Ca(NO_3)_2$,the observed molar mass is $65.4 \, g \, mol^{-1}$ and the theoretical molar mass is $164 \, g \, mol^{-1}$. The degree of dissociation is .............

In a solvent,$50\,\%$ of an acid $HA$ dimerizes and the rest dissociates. The van't Hoff factor of the acid is $.....\times 10^{-2}$. (Round off to the nearest integer)

$A$ $0.5\%$ solution of potassium chloride was found to freeze at $-0.24^{\circ} C$. The percentage dissociation of potassium chloride is .... (Nearest integer)(Molal depression constant for water is $1.80\, K\, kg\, mol^{-1}$ and molar mass of $KCl$ is $74.6\, g\, mol^{-1}$)

$A$ solute $A$ undergoes association in a medium as $nA \rightleftharpoons A_n$. If the degree of association is $\alpha$,then the van't Hoff factor $i$ is given by:

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$A$ $0.5\%$ solution of potassium chloride was found to freeze at $-0.24^{\circ} C$. The percentage dissociation of potassium chloride is .... (Nearest integer)
(Molal depression constant for water is $1.80\, K\, kg\, mol^{-1}$ and molar mass of $KCl$ is $74.6\, g\, mol^{-1}$)