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Graham's law
Concepts tested here
- Rate of effusion and molar mass
All Questions
2012 AIPMT-MAINS 1 question
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A certain gas takes three times as long to effuse out as helium. Its molecular mass will be:Graham's law: rate $\propto \frac{1}{\sqrt M}$, and time taken $\propto \frac{1}{\text{rate}}$, so $\frac{t_{He}}{t_{gas}} = \sqrt{\frac{M_{He}}{M_{gas}}}$
$\frac{1}{3} = \sqrt{\frac{4}{M_{gas}}} \Rightarrow \frac{1}{9} = \frac{4}{M_{gas}}$
$M_{gas} = 4\times9 = 36$ u