1.

The vapour pressure of pure benzene at 88^(@)C is 957mm and that of toluene at the same temperature is 379.5mm. Calculate the composition of a benzene-toluene mixture boiling at 88^(@)C

Answer»

Solution :Since the b.p. is that TEMPERATURE at which the vapour PRESSURE of the liquid is equal to the ATMOSPHERIC pressure, i.e., 760mm.
we have,
`p=x_(A).p_(A)^(0)+x_(B).p_(B)^(0)`…….(Eqn.3)
`760=x_("benzene")957+x_("toluene")379.5`
As `x_("benzene")+x_("toluene")=1`
`:.760=957.x_("benzene")+379.5(1-x_("benzene")`
`:.x_("benzene")=0.6589`


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