$(a)\;36\qquad(b)\;200\qquad(c)\;500\qquad(d)\;18$

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$M=18=\large\frac{Moles\;of\;solute}{Volume\;of\;solution\;in\;litre}$

(i.e) ,18 mole $H_2SO_4$ or $18\times 98g$ $H_2SO_4$ are present in 1000ml solution.

Since density of solution =$1.8g/ml$

Weight of solution =$1.8\times 1000=1800g$

Weight of water =$1800-(18\times 98)$

$\Rightarrow 1800-1764$

$\Rightarrow 36g$

Now ,molality $H_2SO_4=\large\frac{18}{36/1000}$

$\Rightarrow 500$

Hence (c) is the correct answer.

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