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 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

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