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.

Calculate the vapour pressure of solution having 3.42g cane sugar in 1

1.	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.
Answer» Solution :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`