Looking for classes? Ksquare Career Institute, Bengaluru →
A certain gas takes three times as long to effuse out as helium. Its molecular mass will be:
A
9 u
B
27 u
C
36 u
D
64 u
Detailed Solution
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
$\frac{1}{3} = \sqrt{\frac{4}{M_{gas}}} \Rightarrow \frac{1}{9} = \frac{4}{M_{gas}}$
$M_{gas} = 4\times9 = 36$ u