The molality of 90% {H_2SO_4} solution is [de | vedclass

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(C) Given: Density of solution $= 1.8 \ g/mL$,Percentage by mass $= 90\%$.Consider $1 \ L$ $(1000 \ mL)$ of the solution.Mass of solution $= 	ext{Den

Vedclass · Accepted Answer

$9.18$

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(C) Given: Density of solution $= 1.8 \ g/mL$,Percentage by mass $= 90\%$.Consider $1 \ L$ $(1000 \ mL)$ of the solution.Mass of solution $= 	ext{Density} 	imes 	ext{Volume} = 1.8 \ g/mL 	imes 1000 \ mL = 1800 \ g$.Mass of ${H_2SO_4}$ solute $= 90\% 	ext{ of } 1800 \ g = \frac{90}{100} 	imes 1800 = 1620 \ g$.Mass of solvent (water) $= 1800 \ g - 1620 \ g = 180 \ g$.Molar mass of ${H_2SO_4} = (2 	imes 1) + 32 + (4 	imes 16) = 98 \ g/mol$.$	ext{Molality} (m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in } kg} = \frac{1620 / 98}{180 / 1000} = \frac{1620}{98} 	imes \frac{1000}{180} = 9.18 \ m$.

The molality of $90\%$ ${H_2SO_4}$ solution is [density $= 1.8 \ g/mL$].

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