First slide

Gaseous state

Question

Rate of effusion of LPG (a mixture of n-butane and propane) is 1.25 times that of SO₃. Hence, mass fraction of n-butane in LPG is

Moderate

A

0.75
B

0.25
C

0.50
D

0.67

Solution

$\frac{r_{LPG}}{r_{{SO}_{3}}} = \sqrt{\frac{80}{M (LPG)}} = 1 .25 ∴ \frac{80}{M} = (1 .25)^{2} ∴ M (LPG) = 51 .2 M = \frac{M_{1} (butane) X_{1} + M_{2} (propane) X_{2}}{X_{1} + X_{2}} 51 .2 = \frac{58 X_{1} + 44 (1 - X_{1})}{1} X_{1} \approx 0 .50$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App