17.4% (W/V) solution of potassium sulphate ( | vedclass

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(A) For isotonic solutions,the osmotic pressure is equal: $\pi_{1} = \pi_{2}$,which implies $i_{1} C_{1} RT = i_{2} C_{2} RT$.Concentration $C$ in $mo

Vedclass · Accepted Answer

$50$

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(A) For isotonic solutions,the osmotic pressure is equal: $\pi_{1} = \pi_{2}$,which implies $i_{1} C_{1} RT = i_{2} C_{2} RT$.Concentration $C$ in $mol/L$ is calculated as $\frac{10 	imes \% \ (W/V)}{Molar \ Mass}$.For $K_{2}SO_{4}$: $C_{1} = \frac{10 	imes 17.4}{174} = 1 \ M$.For $NaOH$: $C_{2} = \frac{10 	imes 4}{40} = 1 \ M$. Since $NaOH$ is $100\%$ ionised,$i_{2} = 2$ $(NaOH ightarrow Na^{+} + OH^{-})$.Equating the expressions: $i_{1} 	imes 1 = 2 	imes 1 \implies i_{1} = 2$.For $K_{2}SO_{4} ightarrow 2K^{+} + SO_{4}^{2-}$,the number of ions $n = 3$.The degree of ionisation $\alpha$ is given by $\alpha = \frac{i-1}{n-1} = \frac{2-1}{3-1} = \frac{1}{2} = 0.5$ or $50\%$.

$17.4\% \ (W/V)$ solution 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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