Question

The molality of 1 L solution with x% $H_{2}S{O}_4$ is equal to 9. The weight of the solvent present in the solution is 910 g. The value of x in g per 100 mL is :

• A. 90
• B. 80.3
• C. 40.13
• D. 9

Answer 1

The correct option is B 80.3

$Molality=\frac{100\times \text{Weight of solute}}{Mw\times \text{Weight of solvent}}$
$9=\frac{1000\times W}{98\times 910}$
$W=802.6g{L}^{-1}$ $=80.26gper100mL$