Question

# a) Calculate the freezing point of the solution when 1.9 g of MgCl2 (M = 95 g mol−1) was dissolved in 50 g of water, assuming MgCl2 undergoes complete ionization.  (Kf for water = 1.86 K kg mol−1)(b) (i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?(ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution?OR(a)When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (Sx).  (Kf for CS2 = 3.83 K kg mol−1, Atomic mass of sulphur = 32 g mol−1](b)Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing(i)1.2% sodium chloride solution?(ii)0.4% sodium chloride solution?

(i) The elevation in the boiling point of a solution is a colligative property; therefore, it is affected by the number of particles of the solute. Since the amount of solute is higher in 2 M glucose solution as compared to 1 M glucose solution, the elevation in the boiling point is higher. Hence, 2 M glucose solution has a higher boiling point than 1 M glucose solution.

(ii) When the external pressure exerted on the solution is higher than the osmotic pressure, the pure solvent starts flowing out of the solution through the semi- permeable membrane. This process is known as reverse-osmosis.
Or

(b)
(i) 1.2% sodium chloride solution is hypertonic wrt 0.9% sodium chloride solution. Therefore, when the blood cells are placed in 1.2% sodium chloride solution, water flows out of the cells and the cells shrink.

(ii) 0.4% sodium chloride solution is hypotonic wrt 0.9% sodium chloride solution. Therefore, when the blood cells are placed in 0.4% sodium chloride solution, water flows into the cells and the cells swell.