A 1.50 mL sample of a sulphuric acid solution from an automobile storage battery is titrated with 1.47 M odium hydroxide solution to a phenolphthalein endpomt, requiring 23.70 ml. What is the molarity of the sulphuric acid solution?

There is a neutralization reaction happening.

$H_{2}S{O}_4\left(aq\right)+2NaOH\left(aq\right)\to N{a}_{2}S{O}_4\left(aq\right)+2H_{2}O\left(l\right)$

We use the equation provided by the neutralisation reaction to get the acid concentration:

$n_1{M}_1{V}_1=n_2{M}_2{V}_2$

where,
$n_1,M_1$ and ${\mathrm{V}}_1$ are the ${\mathrm{H}}_2{\mathrm{SO}}_4$ acid's n-factor, molarity, and volume.
${\mathrm{n}}_2, {\mathrm{M}}_2 \mathrm{and} {\mathrm{V}}_2$ are the base's $\mathrm{NaOH}$ molarity, n-factor, and volume.

We are given:

$\begin{array}{r}{n}_1=2\\ M_1=?\\ V_1=1.50\mathrm{mL}\\ n_2=1\\ M_2=1.47\mathrm{M}\\ V_2=23.70\mathrm{mL}\end{array}$

Putting values in above equation, we get:

$\begin{array}{r}2\times M_1\times 1.50\mathrm{mL}=1\times 1.47M\times 23.70\mathrm{ml}\\ M_1=\frac{1\times 1.47M\times 23.70\mathrm{ml}}{2\times 1.50\mathrm{ml}}=11.613\mathrm{M}\end{array}$

The molarity of the sulfuric acid solution is 11.613 M.