Question

The molality of pure $H_2{SO}_4$ liquid is:

(Assume density of pure $H_2{SO}_4$ liquid to be $1.25g{ml}^{-1}$)

• A. $10.2$
• B. $12.75$
• C. $1$
• D. $122.5$

Answer 1

The correct option is A $10.2$

Let volume= $1L$, density= $1.25g/ml$

$\therefore$ Mass of $H_{2}S{O}_4=1.25Kg=1250g$

Moles of $H_{2}S{O}_4=\frac{1250}{98}=12.75$

Molality= $\frac{\text{Moles of}{H}_{2}S{O}_4}{\text{Mass of}{H}_{2}S{O}_4\text{(in kg)}}=\frac{12.75}{1.25}=10.2m$