A certain gas effuses through a small opening of a vessel at a rate which is exactly one-fifth the rate at which helium gas effuses under identical conditions. Thus, the molecular weight of the gas is
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Solution:
$\frac{R _{(g)}}{R _{(He)}} = \frac{1}{5} = \frac{4}{M} \Rightarrow M = 100$ .

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A certain gas effuses through a small opening of a vessel at a rate which is exactly one-fifth the rate at which helium gas effuses under identical conditions. Thus, the molecular weight of the gas is

TestHub · Accepted Answer

R ( He ) ​ R ( g ) ​ ​ = 5 1 ​ = M 4 ​ ​ ⇒ M = 100 .

Chemistry - Ideal Gas Question with Solution | TestHub

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