Question

The molality of 90% (mass/volume) $H_2{SO}_4$ solution is:
[Given: density of solution $=1.8$ gm/ml]

• A. 1.8
• B. 48.4
• C. 10.2
• D. 94.6

Answer 1

The correct option is

C

10.2

$90\%m/v{H}_{2}S{O}_4$ means

$90g{H}_{2}S{O}_4$ in $100ml$ of solution

$\Rightarrow 90g{H}_{2}S{O}_4$ in $(100\times 1.8)g$ solution

{$\because$ density of solution $=1.8gm/ml$ }

{ and mass = density $\times$ volume }

Mass of water = 180-90=90g

$\Rightarrow H_{2}S{O}_4$ in $90g$ water

$\therefore$ molality $=\frac{No.ofmolesofsolute}{Massofsolvent\left(Kg\right)}$

$\Rightarrow =\frac{90\times 1000}{98\times 90}$

Molality $=10.204mol/kg$

Molality $=10.204m$

Density of solution is defined as mass per unit volume.