$\begin{array}{1 1}(a)\;537.6ml\qquad(b)\;436ml\qquad(c)\;500ml\qquad(d)\;493.1ml\end{array}$

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Meq of original $H_2SO_4=30\times 1=30$

Meq of $H_2SO_4$ after passing $NH_3=30\times 0.2=6$

Meq of $H_2SO_4$ lost =$30-6=24$

Meq of $NH_3$ passed=Meq of $H_2SO_4$ lost

$\large\frac{w}{17}$$\times 1000=24$

$W_{NH_3}$=0.408g

$\therefore$ Volume of $NH_3$ at STP=$\large\frac{22.4\times 0.408}{17}$

$\Rightarrow 0.5376l$

$\Rightarrow 537.6ml$

Hence (a) is the correct answer.

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