(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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Abnormal molecular mass

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At $300 \ K$,the osmotic pressure of a decinormal solution of sodium chloride is $4.82 \ atm$. The degree of dissociation of sodium chloride is $x \times 10^{-2}$. The value of $x$ is $\left(R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}\right)$

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If $m$ is the normal molecular weight and $x$ is the degree of dissociation of $K_4[Fe(CN)_6]$,then the abnormal molecular weight in the solution of the salt will be $:-$

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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 $= 1.86 \ K \ kg \ mol^{-1}$)

Similar Questions

The molecular weight of benzoic acid dissolved in benzene is found to be ............ $g/mol$.

Which of the following expressions correctly defines the Van't Hoff factor $i$?

$0.004 \ M \ Na_2SO_4$ is isotonic with $0.01 \ M$ glucose. The degree of dissociation of $Na_2SO_4$ is ........... $\%$.

At $300 \ K$,the osmotic pressure of a decinormal solution of sodium chloride is $4.82 \ atm$. The degree of dissociation of sodium chloride is $x \times 10^{-2}$. The value of $x$ is $\left(R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}\right)$

If $m$ is the normal molecular weight and $x$ is the degree of dissociation of $K_4[Fe(CN)_6]$,then the abnormal molecular weight in the solution of the salt will be $:-$

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