# Calculate molality of 1 litre solution of $93\%$ $H_2SO_4$ by volume.The density of solution is $1.84\; g$ $mL^{-1}$

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

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.
edited Mar 18, 2014