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

B

Explanation

Given P= 10 atm,
total numbers of moles, `n_(A)+ n_(B) + n_(C) =10`
`P_(A)=3atm,P_(B)=1atm`
`thereforeP_(A)=x_(A)xxP_("(total)")=(n_(A))/(n_(A)+n_(B)+n_(C))xx10`
`=(n_(A))/10xx10=3`
Similarly, `P_(B)=x_(B)xxP_("(total)")`
So, `n_(B)=1`
`thereforen_(C)=10-(n_(A)+n_(B))=10-4=6`
Mass of C = `6 xx 2` = 12 g