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Pure benzene boiled at 80^(@)C. The boil...

Pure benzene boiled at `80^(@)C`. The boiling point of a solution containing `1 g` of substance dissolved in `83.4 g` of benzene is `80.175^(@)C`. If latent heat of vaporization of benzene is `90 cal per g`, calculated the molecular weight of solute.

Text Solution

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B. pt. of `C_(6)H_(6)=80+273=353 K`,
Latent heat `(l_(v))=90 cal//g`
`DeltaT=80.175-80=0.175, w=1g, W=83.4 g`
`:' K_(b)=(RT^(2))/(1000l_(v))`
`(l_(v) "is latent heat of vaporisation"//g "in the unit of R".)`
or `K_(b)=(2xx353xx353)/(1000xx90)=2.769 K mol^(-1) kg`
Now, `DeltaT_(b)=(K_(b)xx1000xxw)/(m.W)`
`0.175=(2.769xx1000xx1)/(mxx83.4)`
`m=189.72`
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