1.

If we take 44g of CO_(2) and 14g of N_(2) what will be mole fraction of CO_(2) in the mixture

Answer»

`1//5`
`1//3`
`2//3`
`1//4`

Solution :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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