Calculate the vapour pressure of solution having 3.42g cane sugar in 180 g water at `40^(@)C and 100^(@)C` . Given that boiling point of water is `100^(@)C` and heat of vaporization is of `0.2` molal cane solution at `40^(@)C`.

Explanation

At `100^(@)C` , vapour pressure of pure water `(P^(@))=760 mm`
`(P^(@)-P_(s))/(P_(s))=(w_(1)xxm_(2))/(m_(1)xxw_(2))`
`(760-P_(2))/(P_(s))=(3.42xx18)/(342xx180)`
`:.P_(s)=759. 2mm`
`"log" (P_(2))/(P_(T))=(DeltaH_(v))/(2.303R)=[(T_(2)-T_(1))/(T_(1)T_(2))]`
`P_(2)=760 mm, T_(2)= 373K_(f) T_(t)= 313 K`
`Delta=10"Kcal mol"^(-1)`.
`"log"(760)/(P_(1))=(10)/(2.303xx2xx10^(-3))[(373-313)/(373xx313)]`
`:.P_(1)=58.2 mm`
At `40^(@)C,(P^(@)-P_(s))/(P_(s))=(w_(1)xxm_(2))/(m_(2)xxw_(2))`
For `0.2` molal solution `P_(H_(2)O)^(0)=58.2 mm` of Hg at `40^(@)C`
`=(58.2-P_(s))/(P_(s))=(0.2xx18)/(1000)`
`:.P_(s)=57.99 mm`
`:.DeltaO=P^(@)-P_(S)=58.20-57.99`
`n=0.21 mm`