Question

If at a particular temperature, the density of $18M{H}_{2}S{O}_4$ is $1.8$ g cm${}^{-3}$ then calculate the molality of the solution :

Answer 1

Molarity of $H_{2}S{O}_4=18M$

or $18\times 98g{H}_{2}S{O}_4$ is present in 1000 mL solution.

or $1764g{H}_{2}S{O}_4$ is present in $1000\times 1.8=1800gsolution(\because d=1.8gc{m}^{-3})$

$\therefore w_{H_{2}O}=1800-1764=36g$

Now, $Molality=\frac{moleof{H}_{2}S{O}_4}{\text{mass of water}}\times 1000$$=\frac{18\times 1000}{36}=500molal$