1.

A mixture of CO and CO_(2) is found to have a density of 1.5 g/L at 30^(@)C and 740 torr. What is the composition of the mixture.

Answer»

`CO = 0.35775, CO_(2) = 0.64225`
`CO = 0.64225, CO_(2) = 0.3575`
`CO = 0.500, CO_(2) = 0.500`
`CO = 0.2500, CO_(2) = 0.7500`

Solution :`because d=(PM)/(RT)implies M=(dRT)/(P)= (1.5 xx 0.0821 xx 303 xx 76)/(74)=38.276`
Let x MOLE of CO and hence mole of `CO_(2) = (1 – x)`,`38.276 = x xx 28 + (1 – x) 44 implies x = 0.35775`. So `1–x = 0.64225`


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