If 50 % of 1 M Na_2SO_4 is dissociated in a | vedclass

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(A) Given: $1 \ M \ Na_2SO_4$ solution means $1 \ mol$ of $Na_2SO_4$ in $1000 \ mL$ of solution. Density of solution $= 1.2 \ g \ mL^{-1}$. Mass of $1

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$0.95$

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(A) Given: $1 \ M \ Na_2SO_4$ solution means $1 \ mol$ of $Na_2SO_4$ in $1000 \ mL$ of solution. Density of solution $= 1.2 \ g \ mL^{-1}$. Mass of $1000 \ mL$ solution $= 1000 \ mL 	imes 1.2 \ g \ mL^{-1} = 1200 \ g$. Dissociation reaction: $Na_2SO_4 ightarrow 2Na^{+} + SO_4^{2-}$. Since $50 \%$ of $1 \ mol$ of $Na_2SO_4$ dissociates,the amount of $Na_2SO_4$ dissociated $= 0.5 \ mol$. From the stoichiometry,$1 \ mol$ of $Na_2SO_4$ produces $2 \ mol$ of $Na^{+}$,so $0.5 \ mol$ of $Na_2SO_4$ produces $1 \ mol$ of $Na^{+}$. Mass of $Na_2SO_4$ remaining $= (1 - 0.5) 	imes 142 \ g = 71 \ g$. Mass of solvent $= 	ext{Mass of solution} - 	ext{Mass of } Na_2SO_4 = 1200 \ g - 71 \ g = 1129 \ g$. Molality of $Na^{+}$ ion $= \frac{	ext{moles of } Na^{+}}{	ext{mass of solvent in } kg} = \frac{1 \ mol}{1.129 \ kg} \approx 0.885 \ m$. Note: Based on standard calculation,the closest provided option is $0.95 \ m$.

If $50 \%$ of $1 \ M \ Na_2SO_4$ is dissociated in an aqueous solution of density $1.2 \ g \ mL^{-1}$,what is the molality of $Na^{+}$ ion in the solution?

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