Chemistry

Solution - II

Session objectives Colligative properties Relative lowering of vapour pressure Elevation in boiling point Depression in freezing point Osmosis and Osmotic pressure Abnormal molecular mass

Colligative Properties Properties which depend only on the number of particles of solute and donot depend upon the nature of solute particles, (molecules or ions). Relative lowering of vapour pressure. Elevation in boiling point Depression in freezing point Osmotic Pressure

Relative lowering of vapour pressure From Raoult’s law, n moles of solute N moles of solvent

Relative lowering of vapour pressure We can write

Illustrative Example Calculate the relative lowering of vapour pressure of a solution of 6g of urea in 90 g of water. Solution:

Illustrative Example The vapour pressure of pure water at 0°C is 4.579 mm of Hg. A solution of lactose containing 8.45g of lactose in 100 g of water, has a vapour pressure of 4.559 mm at the same temperature. Calculate molecular mass of lactose. Solution :

Illustrative Problem The vapour pressure of pure A is 10 Torr and at the same temperature when 1g B is dissolved in 20 g of A, its vapour pressure is reduced to 9 Torr. If molecular mass of A is 200 amu, calculate the molecular mass of B.

Solution [\ MB = 90 amu]

Elevation in boiling point The boiling point of the solution (TbK) is higher than that of pure solvent (T0 K). A B D C Atmospheric pressure Vapour pressure E F Solvent Solution T0 Tb Temperature

Illustrative Example A solution of 3.8 g of sulphur in 100g of CS2 (Boiling point = 46.30°C) boils at 46.66°C. What is the formula of sulphur molecule in this solution? [ Atomic mass of sulphur = 32 g mol –1 Kb for CS2 = 2.40 K kg mol –1 ]

Solution = 0.36°C = MB = 253 g mol–1 Atomicity = = 8 Hence,Sulphur exists as S8 molecule.

Depression in freezing point The freezing point of a solution (Tf) is less than that of pure solvent (T0). Vapour pressure A B Liquid solvent E D Solution C Solid solvent Tf T T0 Temperature

Osmosis Movement of solvent molecules through a semi permeable membrane from a less concentrated solution to a more concentrated solution.. Semi permeable membrane Allows the passage of solvent molecules but blocks the passage of solute molecules.

Osmotic Pressure The excess hydrostatic pressure on the solution which prevent the movement of solvent molecules through semipermeable membrane. p = Osmotic pressure C = Molar conc. (mol/L) T = Temperature (K) R = Solution constant Isotonic solutions : Equimolar solutions having the same osmotic pressure.

Reverse Osmosis If the pressure higher than the osmotic pressure is applied on the solution, the solvent will flow from the solution into the pure solvent, through the semipermeable membrane Used for water purification.

Illustrative Example A solution containing 8.6 g/l of urea (molecular was - 60 g mol-1) was found to be isotonic with a 5% solution of an organic non-volatile solute at 298 K. What is the molecular mass of the organic solute? Solution :

Illustrative Example What is the relationship between osmotic pressures of 10 grams of glucose(p1),10 grams of urea(p2) and 10 gram of sucrose(p3) which are dissolved in 250 ml of water respectively at 273 K. Solution : Moles of glucose(n1)=10/180 =0.05 Moles of urea(n2)=10/60=0.16 Moles of sucrose(n3)=10/342=0.02 More the number of moles of solute, higher will be the osmotic pressure.

Abnormal molecular mass When the solution is non-ideal i.e. the solution is not dilute. When the solute undergoes association/dissociation . van’t Hoff suggested correction factor (i)

Dissociation of molecule

Association of molecules

Illustrative Example HX H++ X- A 0.2 molal aqueous solution of a weak acid (HX) is 20% ionised. What is the depression in freezing point of the solution (kf for H2O=1.86 K kg mol-1). Solution: HX H++ X- Total no. of moles =

Illustrative Example A solution of x grams of urea in 500 grams of water is cooled to -0.5°C .128 grams of ice separated from the solution. Calculate x, given Kf = 1.86 deg/molal. Solution :

Illustrative Example Among equimolal aqueous solution of C6H5NH2Cl, Ca (NO3)2, La(NO3)2, glucose which will have the highest freezing point? Solution Glucose does not ionise. So the number of particles furnished by glucose in solution will be least compared to other options. Hence, the depression in freezing point is minimum. In other words, the freezing point will be highest for glucose.

Illustrative Example How many grams of NaBr must be added to 270 g of water to lower the vapour pressure by 3.125 mmHg at temperature at which vapour pressure of water is 50 mmHg (Na = 23, Br = 80) ? Assume 100 % ionisation of NaBr.

Solution NaBr taken = 0.5 mol = 0.5 x 103 = 51.5 g

Illustrative Example An aqueous solution contains 5% by mass of urea and 10% by mass of glucose. What will be its freezing point? [Kf for H2O is 1.86 K mol–1 kg]

Solution Five per cent by mass of urea contains 10 per cent by mass of glucose contains \ Total moles of solute = 1.389 = 2.58 \ Freezing point of solution = 0 – 2.58 = –2.58° C or 270.42 K

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