Question

$9.8$ grams of $H_{2}S{O}_4$ is present in two litres of a solution. The molarity of the solution is:

• A. $0.1$ M
• B. $0.05$ M
• C. $0.2$ M
• D. $0.01$ M

Answer 1

Answer: (B)

$0.05$ M

The molar mass of sulphuric acid is $98$ g/mol.

$\text{Number of moles of sulphuric acid}=\frac{\text{Mass of sulphuric acid}}{\text{Molar mass of sulphuric acid}}=\frac{9.8g}{98g/mol}=0.1mol$

$\text{Molarity of the solution}=\frac{\text{Number of moles of sulphuric aicd}}{\text{Volume of solution}}=\frac{0.1mol}{2L}=0.05M$