Question

Rate of diffusion of LPG (mixture of n-butane and propane) is 1.25 times faster than that of $S{O}_3$. Hence, mole fraction of n-butane in LPG is

• A. 0.75
• B. 0.25
• C. 0.87
• D. 0.51

Answer 1

The correct option is D 0.51

$\frac{r_{LPG}}{r_{S{O}_3}}=\sqrt{\frac{M_{S{O}_3}}{M_{LPG}}}=1.25$
$\sqrt{\frac{80}{M_{LPG}}}=1.25$
$\therefore M_{LPG}=\frac{80}{(1.25{)}^2}=51.2$
Mean molar mass of mixture of two gases with mole fractions ${\chi }_1$ and ${\chi }_2$ is
$=M_1{\chi }_1+M_2{\chi }_2$
Gas (1) is n-butane and gas (2) is propane.
Thus, $51.2=58{\chi }_1+44(1-{\chi }_1)$
${\chi }_1=0.51$
Hence, mole fraction of n-butane = 0.51