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Rate of effusion of LPG (a mixture of n- butane and propane) is 1.25 times that of SO3 . What is the mole fraction of n-butane in LPG ?

Pratik , 6 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 6 Years ago

rLPG / rSO3 = √(80/ MLPG) = 1.25 
Therefore, 80/M = (1.25)2 
Therefore, MLPG = 51.2 
M = (M1(butane ) X1 + M2(propane)X2) / (X1 + X2) 
51.2 = 58X1 + 44(1-X1) / 1 
X1 ≈ 0.50
Regards
Arun (askIITians forum expert)

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