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 `:`
A
8 g
B
12 g
C
3 g
D
6 g
Text Solution
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The correct Answer is:
b
(b) Given P = 10 atm, total numbers of moles, `n_(A)+n_(B)+n_(C)=10` `P_(A)=3"atm," P_(B)=1"atm," n_(A)=3,n_(B)=1` `:' P_(A)=x_(A)xxP_(("total"))=(n_(A))/(n_(A)+n_(B)+n_(C))xx10` `=(n_(A))/(10)xx10n_(A)=3` Similarly, `P_(B)=x_(B)xxP_("total")` So, `n_(B)=1` `:. " " n_(C)=10-(n_(A)+n_(B))=10-4=6` Mass of `C=6xx2=12g`
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