Question

Calculate the molarity of solution made by mixing equal volume $60$ by weight of $H_{2}S{O}_4$ having density $1.20$ gm/ml and $70$ by weight of $H_{2}S{O}_4$ having density $1.60$ gm/ml.

Answer 1

Solution 1:- $60\%H_{2}S{O}_4$ by weight, Volume of solution$=\frac{100}{1.2}ml$

i.e. $\frac{100}{1.2}ml$ contains $60g{H}_{2}S{O}_4$

$Vml$ contains $\frac{V\times 60\times 1.2}{100}g{H}_{2}S{O}_4$

Solution 2:- $70\%H_{2}S{O}_4$ by weight, Volume of solution$=\frac{100}{1.6}ml$

i.e. $\frac{100}{1.6}ml$ contains $70g{H}_{2}S{O}_4$

$Vml$ contains $\frac{V\times 70\times 1.6}{100}g{H}_{2}S{O}_4$

Mixing these two, Weight of $H_{2}S{O}_4=\frac{V\times 60\times 1.2}{100}g{H}_{2}S{O}_4+\frac{V\times 70\times 1.6}{100}g{H}_{2}S{O}_4=1.84Vg$

Total Volume$=2Vml$

$\therefore$ Molarity of solution$=\frac{1.84V\times 1000}{98\times 2V}=9.38M$