The $\mathrm{pH}$ of $\mathrm{HCl}$ is 1 . The amount of $\mathrm{NaOH}$ to be added to $100\mathrm{ml}$ of such a $\mathrm{HCl}$ solution to get $\mathrm{pH}$ of 7 is

$\mathrm{pH} \text{ of } \mathrm{HCl}=1$

$\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]$

$-\log \left[{\mathrm{H}}^{+}\right]=1$

$\left[{\mathrm{H}}^{+}\right]=0.1\mathrm{M}$

$\text{ Volume of }\mathrm{HCl}=100\mathrm{ml}$

$\mathrm{HCl}+\mathrm{NaOH}\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}$

According to the equation mole of $\mathrm{HCl}$ reacts with 1 mole $\mathrm{NaOH}$.

Hence$0.1\mathrm{M}$ of $\mathrm{HCl}$ reacts with$0.1\mathrm{M}$ of $\mathrm{NaOH}$.

Amount of $\mathrm{HCl}$ required to prepare$0.1\mathrm{M}$ of $\mathrm{HCl}$ is

$\mathrm{HCl}$

$=0.365 \mathrm{g}$

Amount of $\mathrm{NaOH}$ required to prepare $0.1\mathrm{M}$ of $\mathrm{NaOH}$ is

$0.1\mathrm{M}=\frac{\text{ weight }}{40}\times \frac{1000}{100}$

$=0.4 \mathrm{g}$

Hence the correct answer (B) $0.4 \mathrm{g}$