The molarity of 20% ,(W/W) solution of sulphu | vedclass

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Question

(A) Given: Molarity $(M)$ = $2.55 \, M$,Mass percentage = $20\% \,(W/W)$,Molar mass of $H_2SO_4$ = $98 \, g \, mol^{-1}$.Let the mass of the solution

Vedclass · Accepted Answer

$1.25$

Vedclass · Answer

(A) Given: Molarity $(M)$ = $2.55 \, M$,Mass percentage = $20\% \,(W/W)$,Molar mass of $H_2SO_4$ = $98 \, g \, mol^{-1}$.Let the mass of the solution be $100 \, g$.Mass of solute $(H_2SO_4)$ = $20 \, g$.Number of moles of solute $(n)$ = $\frac{20}{98} \, mol$.Volume of solution $(V)$ = $\frac{	ext{Mass}}{	ext{Density}} = \frac{100}{d} \, mL = \frac{100}{1000d} \, L = \frac{1}{10d} \, L$.Molarity $(M)$ = $\frac{n}{V(L)} = \frac{20/98}{1/10d} = \frac{200d}{98}$.Given $M = 2.55$,so $2.55 = \frac{200d}{98}$.$d = \frac{2.55 	imes 98}{200} = 1.2495 \approx 1.25 \, g \, cm^{-3}$.

The molarity of $20\% \,(W/W)$ solution of sulphuric acid is $2.55 \, M$. The density of the solution is $........... \, g \, cm^{-3}$.

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