Question

What is the strength in $g$ per $litre$ of a solution of sulphuric acid, $12mL$ of which neutralise $15mL$ of $N/10$ sodium hydroxide solution?

Answer 1

Applying $\underset{\left(NaOH\right)}{N_1{V}_1}=\underset{\left(H_{2}S{O}_4\right)}{N_2{V}_2}$
$\frac{1}{10}\times 15=N_2\times 12$
$N_2=\frac{15}{10\times 12}=0.125$
$Normality\times Eq.mass=Strength\left(g/L\right)$
$Strength=0.125\times 49=6.125g/L$.