(A) The dimerization reaction is: $2C_6H_5COOH \to (C_6H_5COOH)_2$Degree of association $\alpha = 0.80$.Van't Hoff factor $i = 1 - \alpha + \frac{\alp

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(A) The dimerization reaction is: $2C_6H_5COOH \to (C_6H_5COOH)_2$Degree of association $\alpha = 0.80$.Van't Hoff factor $i = 1 - \alpha + \frac{\alpha}{n} = 1 - 0.8 + \frac{0.8}{2} = 0.2 + 0.4 = 0.6$.Depression in freezing point formula: $\Delta T_f = i \times K_f \times m$,where $m$ is molality.$m = \frac{w \times 1000}{M_{solute} \times W_{solvent(g)}} = \frac{w \times 1000}{122 \times 30}$.Substituting the values: $2 = 0.6 \times 5 \times \frac{w \times 1000}{122 \times 30}$.$2 = 3 \times \frac{w \times 100}{122 \times 3} = \frac{w \times 100}{122}$.$w = \frac{2 \times 122}{100} = 2.44 \ g$.

Class 12 Chemistry Solutions Abnormal molecular mass

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Molecules of benzoic acid $(C_6H_5COOH)$ dimerise in benzene. '$w$' $g$ of the acid dissolved in $30 \ g$ of benzene shows a depression in freezing point equal to $2 \ K.$ If the percentage association of the acid to form dimer in the solution is $80,$ then $w$ is : ............. $g$
(Given that $K_f = 5 \ K \ kg \ mol^{-1},$ Molar mass of benzoic acid $= 122 \ g \ mol^{-1}$ )

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$2.44$
B
$1.00$
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$1.50$
D
$1.80$

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The observed and calculated molecular mass of silver nitrate $(AgNO_3)$ are $92.64$ and $170$ respectively. The degree of dissociation of silver nitrate is ........ $\%$.

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$HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$
The freezing point depression of a $0.1 \ m$ aqueous solution of a monobasic weak acid $HA$ is $0.20^{\circ} C$. The dissociation constant for the acid is. Given: $K_{f}(H_2O) = 1.8 \ K \ kg \ mol^{-1}$,molality $\equiv$ molarity.