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The vapour pressure of benzene at 90^(@)...

The vapour pressure of benzene at `90^(@)C` is `1020` torr. A solution of `5g` of a solute in `58.5g` benzene has vapour pressure 990 torr. The molecular weight of the solute is?

Text Solution

Verified by Experts

The correct Answer is:
`0.732`
`V_(B) = (78)/(0.877) xx 2750 mL = 244.583 L`
`V_(T) = (92)/(0.867) xx 7720 mL = 819.192 L`
`P_(B) = (1 xx 0.0821 xx293 xx760)/(244.583) = 74.74` torr
`P_(T) = (1 xx 0.0821 xx 293 xx 760)/(819.192) =22.317`
`46 = 74.74 X_(B) +23.317 (1-X_(B))`
`52.423 X_(B) = 23.683`
`X_(B) = 0.451`
`Y_(B) = (1 xx 0.0821 xx 293 xx 760)/(819.192) = 22.317`
`46 = 74.74 X_(B) +23.317 (1-X_(B))`
`54.423 X_(B) = 23.683`
`X_(B) = 0.451`
`Y_(B) = (P_(B)^(@)-X_(B))/(P_(T)) = (74.74xx 0.451)/(46) = 0.732`
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