Question

What is the molarity of $H_{2}S{O}_4$ solution, that has a density $1.84\text{g/cc}$ at $35{}^{\circ }C$ and contains $98\%$ by weight?

• A. $18.4\text{M}$
• B. $18\text{M}$
• C. $4.18\text{M}$
• D. $8.14\text{M}$

Answer 1

The correct option is A $18.4\text{M}$

We know that $98\%$ $H_{2}S{O}_4$ by weight means $98\text{g}$ of $H_{2}S{O}_4$ is present in $100\text{g}$ of solution in water.
Therefore moles of $H_{2}S{O}_4$
$=\frac{\text{Weight of}{H}_{2}S{O}_4}{\text{Molecular weight}}=\frac{98}{98}=1\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100}{1.84}\text{mL}\approx 54.35\text{mL}=\frac{54.35}{1000}\text{L}$
Therefore moalrity of $H_{2}S{O}_4$
$\frac{\text{moles of}{H}_{2}S{O}_4}{\text{Volume (in litres)}}=\frac{1\times 1000}{54.35}=18.399\approx 18.4M$