$(a)\;9.35\qquad(b)\;8.25\qquad(c)\;10.42\qquad(d)\;11.25$

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Given, $H_2SO_4$ is 93% by volume

Weight of $H_2SO_4$ = 93 g

Volume of the solution = 100 mL

Weight of the solution = $100\times 1.84 =1.84\;g$

Weight of water = 184 - 93 = 91 g

Molality=$\large\frac{Mole}{Weight\; of\; water\; in\; kg}$

$\qquad\;\;\;=\large\frac{93}{98\times\Large\frac{91}{1000}}$

$\qquad\;\;\;=10.42$

Hence (c) is the correct answer.

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