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 `:`

Answer

12

Explanation

Given P=10 atm
Total number of moles, `n_(A) + n_(B) + n_(C) =10`
`P_(A) =3 atm, P_(B) = 1 atm`
`therefore P_(A) =x_(A) xx P_("total") =n_(A)/(n_(A) + n_(B) + n_( C)) xx 10`
`=n_(A)/10 xx 10 =3`
So, `n_(B)=1`
`therefore n_( C)=10-(n_(A) + n_(B)) =10-4=6`
Mass of C =`6 xx 2= 12 g`