1.

A mixtureof CO_(2) and CO is passed over red hot graphite when 1 mole of mixture changes to 33.6 L (converted to STP). Hence, mole fraction of CO_(2) in the mixture is

Answer»

0.25
0.33
0.5
0.66

Solution :Suppose `CO_(2)` in the mixture = X MOLE
Then CO in the mixture `=(1-x)` mole
`CO_(2)+Crarr2CO`
Thus, 1 mole of `CO_(2)` gives 2 moles of `CO`
`therefore " x mole of "CO_(2)" will give "CO=2x" mole"`
TOTAL moles of CO after PASSING over RED hot graphite
`=(1-x)+2x=1+x=(1+x)xx22.4L`
Thus, `(1+x)xx22.4=33.6L" (Given)"`
`"or"(1+x)=1.5`
`"or"x=0.5`
Hence, mole fraction of `CO_(2)` in the mixture
`=(0.5)/(1)=0.50`


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