Calculate the molality of 1 L solution of 93 | vedclass

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(D) Given: $93 \%$ $w/V$ $H_2SO_4$ solution means $93 \ g$ of $H_2SO_4$ is present in $100 \ mL$ of solution.For $1 \ L$ $(1000 \ mL)$ of solution, th

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$10.42$

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(D) Given: $93 \%$ $w/V$ $H_2SO_4$ solution means $93 \ g$ of $H_2SO_4$ is present in $100 \ mL$ of solution.For $1 \ L$ $(1000 \ mL)$ of solution, the mass of $H_2SO_4$ solute is $930 \ g$.The density of the solution is $d = 1.84 \ g/mL$.Total mass of $1 \ L$ solution = $Volume 	imes Density = 1000 \ mL 	imes 1.84 \ g/mL = 1840 \ g$.Mass of solvent (water) = $	ext{Total mass of solution} - 	ext{Mass of solute} = 1840 \ g - 930 \ g = 910 \ g = 0.91 \ kg$.Moles of $H_2SO_4$ = $\frac{930 \ g}{98 \ g/mol} \approx 9.49 \ mol$.Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{9.49 \ mol}{0.91 \ kg} \approx 10.42 \ mol/kg$.

Calculate the molality of $1 \ L$ solution of $93 \%$ $H_2SO_4$ by $w/V$ $[d = 1.84 \ g/cc]$.

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