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Rate of effusion of LPG (mixture of buta...

Rate of effusion of LPG (mixture of butane and propane) is 1.25 times that of `SO_3` . Hence, the mole fraction of butane in the mixture is

A

0.75

B

0.25

C

0.5

D

0.67

Text Solution

Verified by Experts

The correct Answer is:
C
`r_(2)//r_(1) = 1.25`
`M_(2) = ? M_(1) = 32 + 16 xx 3 = 80`
`(r_(2))/(r_(1)) = sqrt((M_(1))/(M_(2)))`
`1.25 = sqrt((80)/(M_(2)))`
`M_(2) = (80 xx 16)/(25) = (1280)/(25) = 51.2`
Mol mass of `C_(4)H_(18) = 58`
Mol mass of `C_(3)H_(8) = 44`
Let mole fraction of `C_(4)H_(10) = x`
`:. 58x + 44(1 - x) = 51.2`
`58 x - 44 x = 51.2 - 44.0`
`14 x = 7.2`
`x = (7.2)/(15) ~~ 0.5`
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