a) Calculate the freezing point of the solution when 1.9 g of MgCl (M = 95 g mol) was dissolved in 50 g of water, assuming MgCl undergoes complete ionization. (K for water = 1.86 K kg mol)
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 the solution?
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 the solution?
Solution
(a) Molality:
MgCl Mg + 2Cl, so :
(b)(i) 2 M glucose — elevation of boiling point is colligative, and the more concentrated solution has more solute particles, hence the larger and higher boiling point.
(ii) When the applied pressure exceeds the osmotic pressure, osmosis reverses: pure solvent flows out of the solution through the semipermeable membrane (reverse osmosis, used in water purification/desalination).
MgCl Mg + 2Cl, so :
(b)(i) 2 M glucose — elevation of boiling point is colligative, and the more concentrated solution has more solute particles, hence the larger and higher boiling point.
(ii) When the applied pressure exceeds the osmotic pressure, osmosis reverses: pure solvent flows out of the solution through the semipermeable membrane (reverse osmosis, used in water purification/desalination).
CBSE Class 12 Board · 2016 · Delhi · Subjective