A mixture of CO and `CO_(2)` is found to have a density of `1.50″ g “L^(-1)` at `20^(@)C` and 740 mm pressure. Calculate the composition of the mixture.
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Calculation of average molecular mass of the mixture :
`M=(dRT)/(P)=(1.50″ g “L^(-1)xx0.0821” L atm “mol^(-1)xx293″ K”)/((740//760)” atm”)=37.06`
Calculation of percentage composition :
Suppose mol % of CO in the mixture=x
Then mol% of `CO_(2)` in the mixture=(100-x)
Average molecular mass `=(x xx28+(100-x)xx44)/(100)`
`:. ” ” (28x+4400-44x)/(100)=37.06+” or ” 16x=4400-3706=694″ or ” x=694//16=43.38`
`:.` Mol % CO=43.38 and Mol % of `CO_(2)=100-43.8=56.62`.