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The molality of 1 litre solution with $X\%$ by weight $H_2SO_4$ is equal to 9.The weight of the solvent present in the solution is 910g.The value of $X$ :

$(a)\;90\qquad(b)\;80.3\qquad(c)\;40.13\qquad(d)\;9$

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A)
$m=\large\frac{a\times 1000}{98\times 910}$
$\Rightarrow 9$
$a=802.62g/litre$
$\%$ by weight =80.26g
$\qquad\quad\quad\;\; =X$
Hence (b) is the correct option.
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