A gaseous mixture contains three gaseous `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 and if `C` has molecular weight of `2 g //mol`. Then, the weight of `C` present in the mixture will be `:`

Options

A. 8 g
B. 12 g
C. 3 g
D. 6 g

Answer

b

Explanation

(b) Given P = 10 atm,
total numbers of moles, `n_(A)+n_(B)+n_(C)=10`
`P_(A)=3"atm," P_(B)=1"atm," 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`
Mass of `C=6xx2=12g`