The number of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ions present in $10\mathrm{ml}$ of water at $25°\mathrm{C}$ is

As we know that concentration of ${\mathrm{H}}^{+}$in water is ${10}^{-7}\mathrm{M}$, i.e.,
$\left[{\mathrm{H}}^{+}\right]={10}^{-7} \mathrm{mol}/\mathrm{L}$
Now,
No. of moles of ${\mathrm{H}}^{+}$in 10 $\mathrm{mL}={10}^{-7}\times \frac{10}{1000}={10}^{-9}$
Now as we know that,
No. of ${\mathrm{H}}^{+}$ion in $1 \mathrm{mole}={\mathrm{N}}_{\mathrm{A}}=6.023\times {10}^{23}$
Therefore,
No. of ${\mathrm{H}}^{+}$ion in ${10}^{-9} \mathrm{mole}=6.023\times {10}^{23}\times {10}^{-9}=6.023\times {10}^{14}$.
Hence the correct answer (B) $6.023\times {10}^{14}$.