(a) Define the following terms :
(ii) Molal elevation constant (Kb)
(b) A solution containing 15 g urea (molar mass = 60 g mol–1) per litre of the solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol–1) in water. Calculate the mass of glucose present in one litre of its solution.
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass = 180 g mol–1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution?
(Density of solution = 1.2 g mL–1)
(i) Molarity of a solution is defined as the number of moles of solute present in one litre of the solvent.
(ii) Molal elevation constant (Kb) is defined as the elevation of the boiling point of a solution when one mole of a non-volatile solute is dissolved in one kilogram of a volatile solvent.
(b) Osmotic pressure π = CRT = nVRT/2
where n is the number of moles of solute present in volume V of solvent.
It is given that the solution of urea is isotonic with the solution of glucose, thus
a) A mixture of ethanol and acetone shows positive deviation from Raoult's Law. Pure ethanol possesses hydrogen bonding. The introduction of acetone between the molecules of ethanol results in breaking of some of these hydrogen bonds. Due to the weakening of interactions, the solution shows positive deviation from Raoult’s law.
10% by mass solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 - 10) g = 90 g of water.
Molar mass of glucose (C6H12O6) = 180 g mol-1