Question

Find the strength of 500 mL of a 0.5 N sulphuric acid solution.

• A. 42.5 g/L
• B. 12.25 g/L
• C. 24.5 g/L
• D. 35.68 g/L

Answer 1

The correct option is C 24.5 g/L

$\text{Normality}=\frac{\text{number of gram equivalents}}{\text{total volume of solution (mL)}}\times 1000$
$\Rightarrow 0.5N=\frac{\text{number of gram equivalents}}{500}\times 1000$
Or, number of gram equivalents = 0.25

Mass of sulphuric acid = $0.25\times 49=12.25$ g

Strength of $H_{2}S{O}_4$ solution = $\frac{\text{mass of solute in g}}{\text{total volume of solution in L}}$

Strength = 24.5 g/L