The following is a table of the vapour pressure of pure benzene and chlorobenzene. Determine the boiling point of a mixture containing `40` mole per cent of benzene and `60` mole per cent of chlorobenzene at a pressure of `1000 mm Hg`. `{:(t_(@)C,90,100,110,120,132),("Vapor pressure of benzene"(mmHg),1013,1340,1744,2235,2965),("Vapour pressure of chlorobenzenen"(mm Hg),208,293,403,542,760):}`

Answer

112.4^(@)C`

Explanation

From the datas given in table, we apply roult's law `&` find total pressure obtained by the mixture at difference temperature
At `110^(@)C`
`P_(t)'=X_(A)P_(A)^(@)+X_(B)P_(B)^(@)`
`=0.4xx1744+0.6xx403=939.4 mm Hg`
Similarly at `120^(@)C`,
P_(t)=0.4xx2235 0.6xx542=1217.2mm Hg`
Since the given pressure (i.e. `1000mm Hg`) lies between above two values, So the boiling point of mixturelie between `110^(@)C & 120^(@)C`
Now applying extrapolation method,
"Pressure of mixtue" `{:(t^(@)C,X_(1),X_(2),X_(3)),(,110^(@)C,?,120^(@)C),(,Y_(1),Y_(2),Y_(3)):}`
`939.4 1000 1219.2 mm "of" Hg`.
So `(x_(2)-x_(2))/(x_(3)-x_(1))=(y_(2)-y_(1))/(y_(3)-y_(1))`
`therefore x_(2)=x_(1)=x_(1)+(y_(2)-(y_(1))/(y_(3)-y_(1))xx(x_(3)-x_(1))`
`therefore x_(2)=112^(@)C`