Question

1 litre of a solution of $H_{2}S{O}_4$ contains 392 g of $H_{2}S{O}_4$. The density of the solution is 1.2 gm. $m{l}^{-1}$. Then its molality is:

• A. 5.25 m
• B. 4 m
• C. 4.95 m
• D. 2.95 m

Answer 1

Answer: (C)

4.95 m

Given weight of $H_{2}S{O}_4=392g$

Molecular weight of $H_{2}S{O}_4=98g$

Given volume= $1lit$

Molarity of $H_{2}S{O}_4=\frac{\text{Given weight}}{\text{Given Molecular weight}}\times \frac{1}{V_{\left(lit\right)}}$

$=\frac{392}{98}\times \frac{1}{2}=4$

Density= $1.2g/ml$

Relation between Molarity, Molality & density is

$\frac{1}{Molality}=\frac{Density}{Molarity}-\frac{\text{Molar mass of solute}}{1000}$

$\Rightarrow \frac{1}{Molality}=\frac{1.2}{4}-\frac{98}{1000}=0.3-0.098=0.202$

Molality= $\frac{1}{0.202}=4.95$

$\therefore$ Molality= $4.95$