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A gaseous mixture contains three gases A...

A gaseous mixture contains three gases A,B and C with a total number of moles of 10 and total pressure of 10 atm. The partial pressure of A and B are 3 atm and 1 atm respectively. If C has molecular weight of 2 g/mol. then, the weight of C present in the mixture will be :

A

`8g`

B

`12g`

C

`3g`

D

`6g`

Text Solution

Verified by Experts

The correct Answer is:
2
Given `P=10 atm`
total numbers of moles `,n_(A)+n_(B)+n_(C )=10`
`P_(A)=3atm,P_(B)=1atm,n_(A)=3,n_(B)=1`
`:'P_(A)=x_(A)xxP_(("total"))=(n_(A))/(n_(A)+n_(B)+n_(C)) xx10=(n_(A))/(10)xx10n_(A)=3`
Similarly, `P_(B)=x_(B)xxP_(("total"))`
So,`" "n_(B)=1`
`:. " "n_(C)=10-(n_(A)+n_(B))=10-4=6`
Weight of `C=6xx2=12g`
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