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# Calculate molality of 1L solution of 93% $H_2SO_4$ by volume .The density of solution is 1.84g/ml

$(a)\;11.2\qquad(b)\;9.5\qquad(c)\;8\qquad(d)\;10.42$

$H_2SO_4$ is 93% by volume
Wt of $H_2SO_4=93g$
Volume of solution =100ml
Weight of solution =$100\times 1.84$
$\Rightarrow 184g$
Weight of water =184-93=91g
Molality =$\large\frac{Mole}{Wt \;of\; water\; in\; kg}$
$\Rightarrow \large\frac{93}{98\times\large\frac{91}{1000}}$
$\Rightarrow 10.42$
Hence (d) is the correct option.