(a) The boiling point of benzene is 353.23K. When 1.80g of a non-volatile,non-ionisable solute was dissolved in 90g benzene, the boiling point raised to 354.11K. Calculate molarass of the solute.[Kb for benzene =2.53K kg mol-1].(b)Define :(i) molality of a solution.(ii) Isotonic solutions.

(a) The boiling point of benzene is 353.23K. When 1.80g of a non-volat

1.	(a) The boiling point of benzene is 353.23K. When 1.80g of a non-volatile,non-ionisable solute was dissolved in 90g benzene, the boiling point raised to 354.11K. Calculate molarass of the solute.[Kb for benzene =2.53K kg mol-1].(b)Define :(i) molality of a solution.(ii) Isotonic solutions.
Answer» Solution :(a) Molality: Weight of KI in 100 g of water=20g weight of water in the solution=100-20=80=0.08kg Molar mass of KI =39+127=166`mol^(-1)` Molality of the solution (m) Number of MOLES of KI/Mass of water in Kg`1.506.mol .kg^(-1)=1.506m` Molality : weight of the solution =100G Density of the solution =`1.202 g mL^-1` Volume of the solution =weight of the solution/Density ` (100g)/(1.202 g mL^(-1))` `=83.19 mL =0.083 L` Molarity of the solution (M) =`(20g)/((166 g mol^(-1))` `1.45 mol L^(-2)=1.45M` (C) Moles of FRACTION of KI: (Number of moles of KI) `n_(KI)`=Mass of KI/Molar mass of KI `((20g))/((166 g mol^-1))=0.12 mol` (Number of moles of water `nH_2O`=Mass of water/Molar mass of water =`((80 g))/((18 g mol^(-1)))=4.44 mol` Mole if fraction of KI `x_(KI)=n_(KI)/(n_(KI)+n_(H_2O))=((0.02mol))/((0.12+4.44)mol)=0.12/4.56=0.0263` Two different solutions having the same osmotic pressure are called isotonic solutions. (b) (i) It is defined as number of moles of the solute dissolved in one kilogram (1000g) of the solvent. (ii) Two solutions having same osmotic pressure at a given temperay are called isotonic solutions.