What is the freezing point of 0.1 m NaCl?

For 0.1 m NaCl:

NaCl   →   Na+     +    Cl-

0.1 m       0.1 m         0.1 m 

Total particles in solution = 0.2 mol

For 0.05 m Al2(SO4)3:

Al2(SO4)3    →  2Al3+     + 3`"SO"_4^(2-)`

0.05 m             0.1 m            0.15 m

Total particles in solution = 0.25 mol

Al2(SO4)3 solution contains more number of particles than NaCl solution. Hence, Al2(SO4)3 solution has maximum ΔTf.

Therefore, the freezing point depression of 0.05 m Al2(SO4)3 solution will be higher than 0.1 m NaCl solution.

Answer

Vapour pressure of pure water at 298 K is 23.8 mm. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Vapoure pressure of pure water (solvent) at 298 K, p0 = 23.8 mmVapour pressure of solution, p = ?Mass of solvent ,W = 850 gMass of solute,M = 50 g

Mol. mass of water (H2O), M = 18 g mol–1
Mol.mass of urea NH2 CO NH2
= 14 + 2 + 12 + 16 + 14 + 2

                     p=p0-w×Mm×W×p°

                   p=23.8-50×1860×850

                      =23.8-0.017=23.78