The molar volume of liquid benzene (density =0.877 g ml^(-1)) increases by a factor of 2750 of it vaporises at 20^(@)C At 27^(@)C when a non-votalite solute (that does not to be 98.88 mm Hg. Calculate the freezing point of the solution. Given Enthalphy f vaporization of benzene (f) 394.57 J g^(-1) Enthalpy of fusion constant for benzne =5.0 kg "mol"^(-1)

The molar volume of liquid benzene (density =0.877 g ml^(-1)) increase

1.	The molar volume of liquid benzene (density =0.877 g ml^(-1)) increases by a factor of 2750 of it vaporises at 20^(@)C At 27^(@)C when a non-votalite solute (that does not to be 98.88 mm Hg. Calculate the freezing point of the solution. Given Enthalphy f vaporization of benzene (f) 394.57 J g^(-1) Enthalpy of fusion constant for benzne =5.0 kg "mol"^(-1)
Answer» SOLUTION :LET moles of benzene vaproises, at `20^(@)C=n_(1)` volume of `n_(1)` mole of benzene `(f)=(78n_(1))/(0.877)` Volume of `n_(1)` mole of benzene `(g)=2750 ((78n_(1))/(0.877))` `PV=nRT` `P^(@)=((78n_(1))/(0.877))((2750)/(1000))=n_(1)xx0.082xx293` `P_("Benzene")=0.0982` atm. `=74.63 mm Hg` `P_("Benzene")^(@)` at `27^(@)C` can be CALCULATED as `log((P_(2))/(P_(1)))=(DeltaH_("vap"))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `log ((P_(2))/(74.63))=(394.57xx78)/(2.303xx8.314)[(7)/(300xx293)]` `P_("Benzene")^(@) ("at"27^(@)C)=100.2` mm Hg Molality of the solution `=(X_("solvent")xx1000)/(X_("solvent")xx78)=(((100.2-98.88)/(98.88))xx10000)/(0.98xx78)=0.17` `Deltat_(f)=K_(f)m=5xx0.17=0.85` We known that `K_(f), =[(RT_(f)^(2))/(1000DeltaH_(f))]M` `5=((8.314xxT_(1)^(2))/(1000xx10060))78, T_(f)-278.5K`