Calculate the molality of 1 L solution of 93 | vedclass

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(D) Given: Volume of solution = $1 \ L = 1000 \ mL$. Concentration = $93 \ \%$ (w/v) $H_2SO_4$,which means $93 \ g$ of $H_2SO_4$ is present in $100 \

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$10.43$

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(D) Given: Volume of solution = $1 \ L = 1000 \ mL$. Concentration = $93 \ \%$ (w/v) $H_2SO_4$,which means $93 \ g$ of $H_2SO_4$ is present in $100 \ mL$ of solution. Therefore,mass of $H_2SO_4$ in $1000 \ mL$ = $930 \ g$. Molar mass of $H_2SO_4$ = $98 \ g/mol$. Moles of $H_2SO_4$ = $\frac{930}{98} \approx 9.49 \ mol$. Mass of solution = $	ext{Density} 	imes 	ext{Volume} = 1.84 \ g/mL 	imes 1000 \ mL = 1840 \ g$. Mass of solvent (water) = $	ext{Mass of solution} - 	ext{Mass of solute} = 1840 \ g - 930 \ g = 910 \ g = 0.91 \ kg$. Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{9.49}{0.91} \approx 10.43 \ m$.

Calculate the molality of $1 \ L$ solution of $93 \ \%$ $H_2SO_4$ (weight/volume). The density of the solution is $1.84 \ g/mL$.

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