The density (in g mL^{-1}) of a 3.60 M sulp | vedclass

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(C) Given that,mass $\%$ of $H_2SO_4 = 29 \%$. $100 \ g$ of solution contains $29 \ g$ of $H_2SO_4$. Let the density of the solution be $d \ g \ mL^{-

Vedclass · Accepted Answer

$1.22$

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(C) Given that,mass $\%$ of $H_2SO_4 = 29 \%$. $100 \ g$ of solution contains $29 \ g$ of $H_2SO_4$. Let the density of the solution be $d \ g \ mL^{-1}$. Volume of $100 \ g$ solution $= \frac{	ext{mass}}{	ext{density}} = \frac{100}{d} \ mL$. Moles of $H_2SO_4 = \frac{29 \ g}{98 \ g \ mol^{-1}} \approx 0.2959 \ mol$. Molarity $(M) = \frac{	ext{moles of solute}}{	ext{volume of solution in } L} = \frac{0.2959 \ mol}{(100/d) 	imes 10^{-3} \ L} = 3.60 \ M$. $\frac{0.2959 	imes d 	imes 1000}{100} = 3.60$. $2.959 	imes d = 3.60$. $d = \frac{3.60}{2.959} \approx 1.216 \ g \ mL^{-1}$,which is approximately $1.22 \ g \ mL^{-1}$.

The density (in $g \ mL^{-1}$) of a $3.60 \ M$ sulphuric acid solution that is $29 \% \ H_2SO_4$ (Molar mass $= 98 \ g \ mol^{-1}$) by mass will be

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