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. `8g`
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`
`therefore 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= 6xx 2=12g`.