The equivalent weight of ${\mathrm{CH}}_4$ in the reaction ${\mathrm{CH}}_4+2{\mathrm{O}}_2\to {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$ is $[\mathrm{M}=$ mol. wt ]

we know that equivalent weight$=\frac{\mathrm{Molecular} \mathrm{Weight}}{\mathrm{Net} \mathrm{positive} \mathrm{valence}}$

In the reaction

${\mathrm{CH}}_4+2{\mathrm{O}}_2\to {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$

The net valence of ${\mathrm{CH}}_4$=8

Therefore, equivalent weight of ${\mathrm{CH}}_4$$=\frac{\mathrm{M}}{8}$

Hence, option(b) is the answer