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

Answer

2

Explanation

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`