Under Graham's law of effusion, if an unknown gas effuses 1.414 times faster than sulfur dioxide gas (SO2, molar mass = 64 g/mol), what is the molar mass of the unknown gas?
Answer and explanation
Correct answer: B. 32 g/mol
- Step 1: Graham's law: r1 / r2 = sqrt(M2 / M1). Step 2. r_unknown / r_SO2 = 1.414 = sqrt(2). Step 3. sqrt(64 / M_unknown) = sqrt(2) => 64 / M_unknown = 2 => M_unknown = 64 / 2 = 32 g/mol (such as O2). Answer: 32 g/mol. Why not the others: 16 g/mol would effuse 2.0 times faster; 128 g/mol would effuse slower.
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