Question

A piece of aluminium weighing $2.7g$ is heated with $75mL$ of $H_{2}S{O}_4$ (sp. gr. $1.18$, containing $24.7$% $H_{2}S{O}_4$ by mass). After the metal is carefully dissolved, the solution is diluted to $400mL$. Calculate the molarity of the free $H_{2}S{O}_4$ in the resulting solution.

Answer 1

Mass of $H_{2}S{O}_4=\frac{24.7}{100}\times 75\times 1.18$
$=21.8595g$
Reaction: $\underset{2\times 27}{2Al}+\underset{3\times 98}{3H_{2}S{O}_4}\to A{l}_2(S{O}_4{)}_3+3H_2$
$H_{2}S{O}_3$ required for dissolving $2.7gAl$
$=\frac{3\times 98}{2\times 27}\times 2.7=14.7g$
$H_{2}S{O}_4$ left unreacted $=(21.985-14.7)g=7.1595g$
$7.1595g{H}_{2}S{O}_4$ is present in $400mL$
Amt. of $H_{2}S{O}_4$ present in one litre $=\frac{7.1595}{400}\times 1000g$
$=17.898g$
No. of $g$ moles of $H_{2}S{O}_4=\frac{17.898}{98}=0.1826$
Hence, molarity of $H_{2}S{O}_4=0.1826M$.