Question

Rate of effusion of LPG (a mixture of n-butanemole fraction of n-butane in LPG is

• A. 0.75
• B. 0.25
• C. 0.5
• D. 0.67

Answer 1

Answer: (C)

0.5

Answer 2

$r_2/r_1$ = 1.25 $M_2 = ? . M_1$ = 32 + 16 $\times$ 3 = 80 $\frac{r_2}{r_1}= \sqrt{\frac{M_1}{M_2}}$ ; 1.25 = $\sqrt{\frac{80}{M_2}}$ $M_2 = \frac{80 \times 16}{25} = \frac{1280}{25}$ =51.2 Mol mass of $C_4H_{18}$ = 58 Mol mass of $C_3H_8$ = 44 Let mole fraction of $C_4H_{10}$ = x $\therefore$ 58x + 44(1 -x ) = 51.2 58r - 44x = 51.2 - 44.0 14x = 7.2 $x = \frac{7.2}{15} = 0.5$