1. A gas contains 50% helium and 50% methane by volume. What is the percentage
by weight of methane in the mixture?
a. 25%
b. 40%
c. 50%
d. 80%
A gas mixture with 50% helium and 50% methane by volume has 80% methane by weight due to methane’s higher molecular mass. This calculation hinges on equal moles from equal volumes per Avogadro’s law, shifting weight balance toward methane.
Step-by-Step Solution
Equal volumes mean equal moles: 1 mole helium (4 g/mol) and 1 mole methane (16 g/mol). Total mass equals 4 g + 16 g = 20 g. Methane weight percent is (16/20) × 100 = 80%.
Average molecular weight of mixture is (4 + 16)/2 = 10 g/mol. Methane fraction by weight uses its contribution: (16/20) × 100 = 80%, confirming the result.
Option Analysis
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a. 25%: Incorrect; assumes equal weights from volumes, ignoring helium’s lower mass leads to underestimating methane.
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b. 40%: Wrong; might stem from averaging masses without mole equality, like (4+16)/40 error.
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c. 50%: Fails to convert volume percent to weight; treats percentages as interchangeable.
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d. 80%: Correct; properly accounts for mole-to-mass conversion in gas mixtures.
Exam Relevance
This problem tests gas laws and stoichiometry, common in CSIR NET Life Sciences (biochemistry interfaces) and chemistry exams. Practice distinguishes volume from weight percents for mixtures like noble gases with hydrocarbons.
