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A gaseous mixture contains three gases A...

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. If C has molecular weight of 2 g/mol. then, the weight of C present in the mixture will be :

A

8

B

12

C

3

D

6

Text Solution

Verified by Experts

The correct Answer is:
B

`P = 10 atm, n_(A) + n_(B) + n_(C) = 10`
`p_(A) = 3.0 atm, p_(B) = 1.00 atm`
`p_(A) = underset(n_(A) + n_(B) + n_(C))overset(n_(A))rarr P`
`3.00 = (n_(A))/(10) 10 rArr n_(A) = 3.00`
Similarly, `n_(B) = 1.00`
`n_(C) = (n_(A) + n_(B) + n_(C)) - (n_(A) + n_(B))`
`= 10-(3 + 1) = 6`
Mass of gas `C = n_(C) xx` Mol. mass of C
`= 6 xx 2 = 12`
`n = (2.8)/(28) = 0.1`
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