# A piece of Al weighing $2.7g$ is titrated with $75\;ml$ of $H_2SO_4$ (specific gravity = $1.18gmL^{-1}$ and $24.7%$ $H_2SO_4$ by weight). After the metal is completely dissolved, the solution is diluted to $400\;ml$. Calculate molarity of free $H_2SO_4$ in solution.

$(a)\;0.214\qquad(b)\;0.183\qquad(c)\;0.32\qquad(d)\;0.225$

Weight of Al = 2.7g
Equivalent weight of Al = $\large\frac{2.7}{9}$$=0.3 Molecular equivalent weight of Al = 0.3\times 1000=300 \therefore Weight of H_2SO_4 = 24.7g Weight of the solution = 100g \therefore Volume of the solution = \large\frac{100}{1.18}$$ml=84.75ml$
$N_{H_2SO_4}=\large\frac{24.7}{49\times \Large\frac{100}{1.18\times 1000}}=$ $5.95$
Now, molecular equivalent weight of $H_2SO_4$ in 75ml = $5.95\times 75=446.25$
Molecular equivalent weight of Al added = 300
$\therefore$ Molecular equivalent weight of $H_2SO_4$ left after the reaction = $446.25-300 =146.25$
After the metal is completely dissolved, the solution is diluted to 400ml.
$\therefore N_{H_2SO_4}$ left = $\large\frac{146.25}{400}$$=0.367 \therefore M_{H_2SO_4} left = \large\frac{0.367}{2}$$=0.183$
Hence (b) is the correct answer.
edited Mar 18, 2014