17.4% text{ (mass/volume) } K_2SO_4 solution | vedclass

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(C) Molar mass of $K_2SO_4 = 2 	imes 39 + 32 + 4 	imes 16 = 174 \ g/mol$. Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$. Molarity of $K_2SO_4$ $(C

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$50$

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(C) Molar mass of $K_2SO_4 = 2 	imes 39 + 32 + 4 	imes 16 = 174 \ g/mol$. Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$. Molarity of $K_2SO_4$ $(C_1) = \frac{17.4 	imes 1000}{174 	imes 100} = 1 \ M$. Molarity of $NaCl$ $(C_2) = \frac{5.85 	imes 1000}{58.5 	imes 100} = 1 \ M$. For isotonic solutions,$i_1 C_1 = i_2 C_2$. For $NaCl$,$i_2 = 2$ (as it is $100\%$ ionized). $i_1 	imes 1 = 2 	imes 1 \Rightarrow i_1 = 2$. For $K_2SO_4$,$i_1 = 1 + (n-1)\alpha = 1 + (3-1)\alpha = 1 + 2\alpha$. $1 + 2\alpha = 2$ $\Rightarrow 2\alpha = 1$ $\Rightarrow \alpha = 0.5$. $\%$ ionization $= 0.5 	imes 100 = 50\%$.

$17.4\% \text{ (mass/volume) } K_2SO_4$ solution at $27^\circ C$ is isotonic to $5.85\% \text{ (mass/volume) } NaCl$ solution at $27^\circ C$. If $NaCl$ is $100\%$ ionized,what is the $\%$ ionization of $K_2SO_4$ in aqueous solution? [Atomic weights: $K = 39, Na = 23, S = 32, O = 16, Cl = 35.5$]

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