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

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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.

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