17.4% (W/V) of potassium sulphate (mol. wt. = | vedclass

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(A) For $K_{2}SO_{4}$:$K_{2}SO_{4} \longrightarrow 2K^{+} + SO_{4}^{2-}$At equilibrium,the van't Hoff factor $(i_{1}) = 1 + 2\alpha$,where $\alpha$ is

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) For $K_{2}SO_{4}$:$K_{2}SO_{4} \longrightarrow 2K^{+} + SO_{4}^{2-}$At equilibrium,the van't Hoff factor $(i_{1}) = 1 + 2\alpha$,where $\alpha$ is the degree of dissociation.For $NaOH$:$NaOH \longrightarrow Na^{+} + OH^{-}$Since $NaOH$ is $100\%$ ionised,$i_{2} = 2$.Molar concentration of $K_{2}SO_{4} (C_{1}) = \frac{17.4 \ g / 174 \ g/mol}{0.1 \ L} = 1 \ M$.Molar concentration of $NaOH (C_{2}) = \frac{4 \ g / 40 \ g/mol}{0.1 \ L} = 1 \ M$.Since the solutions are isotonic,$\pi_{1} = \pi_{2}$,so $i_{1}C_{1}RT = i_{2}C_{2}RT$.$(1 + 2\alpha) 	imes 1 = 2 	imes 1$.$1 + 2\alpha = 2$ $\Rightarrow 2\alpha = 1$ $\Rightarrow \alpha = 0.5$.Thus,the degree of ionisation is $50 \%$.

$17.4\% (W/V)$ of potassium sulphate $(mol. wt. = 174)$ is isotonic with $4\% (W/V)$ aqueous solution of $NaOH$. If $NaOH$ is $100\%$ ionised,the degree of ionisation of potassium sulphate is $.......... \%$.

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