The strength of ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution (in gm/ litre), 12 mL of which neutralized by 15 mL of $\frac{\mathrm{M}}{10}$ $\mathrm{NaOH}$ solution :

$\underset{\left(\mathrm{NaOH}\right)}{{\mathrm{N}}_1{\mathrm{V}}_1}=\underset{\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)}{{\mathrm{N}}_2{\mathrm{V}}_2}$

$\frac{1}{10}\times 15=N_2\times 12$

$N_2=\frac{15}{10\times 12}=0.125$

$\text{ Normality }\times \text{ Eq.mass }=\text{ Strength }(\mathrm{g}/\mathrm{L})$

$\text{ Strength }=0.125\times 49=6.125\mathrm{g}/\mathrm{L}\text{. }$