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Molarity of $H_2SO_4$(density 1.8g/ml) is 18m.The molality of this $H_2SO_4$ is

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

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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.
answered Oct 29, 2013 by sreemathi.v
 

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