Under identical conditions a gas X diffuses at 0.40 times the rate of CH4 (molar mass 16). The molar mass of X in g mol^-1 (one decimal) is

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Under identical conditions a gas X diffuses at $0.40$ times the rate of $CH X_{4}$ (molar mass 16). The molar mass of X in $g mo l^{- 1}$ (one decimal) is

Numerical answer type, answer in g/mol, tolerance 0.1.

Graham's law:

\frac{r _{X}}{r _{CH X_{4}}} = \frac{16}{M _{X}} = 0.40

M_{X} = \frac{16}{0.16} = 100.0 g mo l^{- 1}