Question

The rate of effusion of LPG (a mixture of n-butane and propane) is 1.25 times that of $S{O}_3$. Hence, mole fraction of n-butane in LPG is .

• A. 0.5
• B. 0.25
• C. 0.75

Answer 1

The correct option is A 0.5

$\frac{r_{LPG}}{r_{S{O}_3}}=\sqrt{\frac{80}{M_{LPG}}}=1.25\therefore \frac{80}{M}=⟮1.25{⟯}^2$

$\therefore M_{LPG}=51.2$

$M=\frac{⟮M_1⟮butane⟯X_1+M_2⟮propane⟯X_2⟯}{X_1+X_2}$

$51.2=\frac{58X_1+44⟮1-X_1⟯}{1}$

$X_1\approx 0.50$