Question

Commercially available $H_{2}S{O}_4$ is $98$gms by weight of $H_{2}S{O}_4$ and $2$gms by weight of water. It's density is $1.83$g $c{m}^{-3}$. Calculate the molality (m) of $H_{2}S{O}_4$ solution.

[Take : Molar mass of $H_{2}S{O}_4$ as $98$g $mo{l}^{-1}$].

• A. $500$ m
• B. $20$ m
• C. $50$ m
• D. $200$ m

Answer 1

Answer: (A)

$500$ m

Given that

Commercially available $H_{2}S{O}_4$ is $98$ gms by weight of $H_{2}S{O}_4$ and $2$ gms by weight of water.

It's density is $1.83$$g$. $c{m}^{-3}$.

The molar mass of $H_{2}S{O}_4$ is $98$ $g$. $mo{l}^{-1}$.

Now,

Molality $=\frac{\text{weight (in g)}of{H}_{2}S{O}_4}{\text{molecular weight of}{H}_{2}S{O}_4}\times \frac{\text{1000 g/kg}}{\text{weight of water (in g)}}$

$=\frac{\text{98 g}}{\text{98 g/mol}}\times \frac{\text{1000 g/kg}}{\text{2 g}}=500$m.

Therefore, molality (m) of $H_{2}S{O}_4$ is $500$m.