The $\mathrm{pH}$ of a sample of ${\mathrm{H}}_2{\mathrm{SO}}_4$ is $1.3979$. The percentage of the solution is $73.5\%(\mathrm{w}/\mathrm{w})$, the density of the solution is

Normality$=\frac{d\times \%w/\omega \times 10}{\text{ equo nit. }}$

$\text{ guien }-\log \left[H^{+}\right]$$=1.3979$

$\left[{\mathrm{H}}^{+}\right]$ =4 $\times {10}^{-2}$

Then Normality $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)=4\times {10}^{-2}$.

So, $4\times {10}^{-2}=\frac{d\times 73.5\times 10}{49}$

Then $d=2.66\times 10.3 \mathrm{g}/\mathrm{cc}$.

Hence the correct answer is  (A) $2.66\times {10}^{-3} \mathrm{g}/\mathrm{cc}$