17.4%, (w/v) K_2SO_4 (Mw=174, g/mol) solution | vedclass

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(D) Two solutions are isotonic if their molar concentrations multiplied by their van't Hoff factors $(i)$ are equal: $(i 	imes C)_{K_2SO_4} = (i 	im

Vedclass · Accepted Answer

$50$

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(D) Two solutions are isotonic if their molar concentrations multiplied by their van't Hoff factors $(i)$ are equal: $(i 	imes C)_{K_2SO_4} = (i 	imes C)_{NaOH}$.For $NaOH$ $(Mw=40\, g/mol)$: $C = \frac{4\, g}{40\, g/mol 	imes 0.1\, L} = 1\, M$. Since $NaOH$ is $100\%$ ionized,$i = 2$.For $K_2SO_4$ $(Mw=174\, g/mol)$: $C = \frac{17.4\, g}{174\, g/mol 	imes 0.1\, L} = 1\, M$.Equating the two: $i_{K_2SO_4} 	imes 1 = 2 	imes 1$,so $i_{K_2SO_4} = 2$.For $K_2SO_4 ightarrow 2K^{+} + SO_4^{2-}$,the number of ions $n = 3$.Using the formula $i = 1 + \alpha(n - 1)$:$2 = 1 + \alpha(3 - 1)$$2 = 1 + 2\alpha$$1 = 2\alpha$$\alpha = 0.5$ or $50\%$.

$17.4\%\, (w/v)$ $K_2SO_4$ $(Mw=174\, g/mol)$ solution at $27\, ^\circ C$ is isotonic to $4\%\, (w/v)\, NaOH$ solution at the same temperature. If $NaOH$ is $100\%$ ionized,what is the $\%$ ionization of $K_2SO_4$ in aqueous solution .......... $\%$.

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