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A gas mixture 44g of CO(2) and 14g of N(...

A gas mixture 44g of `CO_(2)` and 14g of `N_(2)`, what will be fraction of `CO_(2)` in the mixture

A

`1//5`

B

`1//3`

C

`2//3`

D

`1//4`

Text Solution

Verified by Experts

The correct Answer is:
C
Mole fraction of `CO_(2)=(n_(CO_(2)))/(n_(CO_(2))+n_(N_(2)))=((44)/(44))/((44)/(44)+(14)/(28))=(2)/(3)`.
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