An equimolar mixture of $He$ (molar mass 4) and $CH X_{4}$ (molar mass 16) effuses through a pinhole. In the initial stage, when the composition inside has not changed appreciably, $1.00 mmol$ of gas has effused in total. The amount of $He$ in this effused gas in mmol (two decimals) is

Numerical answer type, answer in mmol, tolerance 0.01.

Answer: 0.67 mmol

Effusion rate of each gas

\propto p_{i} / M_{i}

; with equal partial pressures,

\frac{n _{He}}{n _{CH X_{4}}} = \frac{16}{4} = 2

. So the effused gas is

2/3

He:

1.00 \times \frac{2}{3} = 0.667 \approx 0.67 mmol