An aqueous solution of 98% (w/w) H_2SO_4 has | vedclass

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(B) Given: $98\% (w/w)$ $H_2SO_4$ means $98 \ g$ of $H_2SO_4$ in $100 \ g$ of solution. Mass of solute $(H_2SO_4) = 98 \ g$. Molar mass of $H_2SO_4 =

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$500$

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(B) Given: $98\% (w/w)$ $H_2SO_4$ means $98 \ g$ of $H_2SO_4$ in $100 \ g$ of solution. Mass of solute $(H_2SO_4) = 98 \ g$. Molar mass of $H_2SO_4 = 2 	imes 1 + 32 + 4 	imes 16 = 98 \ g/mol$. Moles of $H_2SO_4 = \frac{98 \ g}{98 \ g/mol} = 1 \ mol$. Mass of solvent (water) = Total mass - Mass of solute = $100 \ g - 98 \ g = 2 \ g = 0.002 \ kg$. Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{1 \ mol}{0.002 \ kg} = 500 \ m$.

An aqueous solution of $98\% (w/w)$ $H_2SO_4$ has a density of $1.84 \ g/mL$. The molality of the solution is: (in $m$)

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