A gaseous mixture of three gases A, B and C has a pressure of 10 atm. The total number of moles of all the gases is 10. If the partial pressure of A and B are 3.0 and 1.0 atm respectively and if C has mol. wt. of 2.0, what is the weight of C in g present in the mixture ?
Text Solution
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Pressure of Total mixture = 10 atm
P A + P B + P C = 10
3 + 1 + P C = 10 ⇒ P C = 6 atm
Total moles of mixture = 10
n A + n B + n C = 10
⇒ 
Let n A = K ⇒ n B = 
n C = 
n B = 2K
⇒ K + 
+ 2K = 10 ⇒ 
= 
⇒ n C = 2K
⇒ 
= 10 K = 3, ⇒ n A = 3
n B = 1
n C = 6
weight of 'C' in mixture = 2 × 6 = 12.
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