A sealed rigid container holds an equimolar mixture of Helium () and Methane (). If a tiny pinhole i — States of Matter and Gaseous State Chemistry Question
Question
A sealed rigid container holds an equimolar mixture of Helium ($He$) and Methane ($CH_4$). If a tiny pinhole is made, what is the exact initial composition by mass of the effusing gas mixture (He : CH4)?
Answer: B
💡 Solution & Explanation
Initial mole ratio is $1:1$, so partial pressures are equal. Rate ratio (moles) $r_{He}/r_{CH4} = \sqrt{16/4} = 2/1$. The effusing gas contains $2\text{ moles}$ of $He$ for every $1\text{ mole}$ of $CH_4$. Mass ratio = $(2\text{ mol} \times 4\text{ g/mol}) / (1\text{ mol} \times 16\text{ g/mol}) = 8 / 16 = 1:2$.
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