Question

A storage battery contains a solution of $H_{2}S{O}_4$ 30% by weight. Find out molality of solution?
(Given, density of solution $=1.2g/c{m}^3$)

• A. 2.07 m
• B. 6.07 m
• C. 4.37 m
• D. 3.37 m

Answer 1

The correct option is C 4.37 m

Concentration of $H_{2}S{O}_4$ = 30% by weight
Weight percent is the weight in grams of the solute per 100 g of the solution.​
$\Rightarrow W_{\text{solute}}=30g\Rightarrow W_{\text{solution}}=100g\Rightarrow W_{\text{solvent}}=70g$

Molar mass of $H_{2}S{O}_4$ = 98 $gmo{l}^{-1}$
Weight of solute, $W_{\text{solute}}=30g$,
$Numberofmolesofsolute=\frac{Mass}{Molarmass}$
$n_{\text{solute}}=\frac{30}{98}=0.306$
$W_{\text{solvent}}=70g$

Molality is the number of moles of the solute per kg of the solvent or per 1000 grams of the solvent.​

$\text{Molality (m)}=\frac{n_{\text{solute}}\times 1000}{W_{\text{solvent}}ingram}$
$m=\frac{0.306\times 1000}{70}=4.37m$