Question

calculate the new freezing point of water when 72.0 g of BaCl2 are added to 0.175 kg of water

Answer 1

1. Find the molality of the solution.

molality is the ratio of the amount of solute (mol) to the mass of solvent (kg):

m= n(BaCl₂) / 0,175kg = 0,346/0,175 =1,977

[ n(BaCl₂) = 72 / 208 = 0,346 mol ]

2. Now substitute this value into the freezing point expression and solve for ∆T:

∆T = i * m * K_f

i – van't Hoff Factor= 3 (BaCl₂ =>Ba²⁺ + 2Cl^-

dissociation produces 3 ions);

K_f — freezing point constant = 1.86

∆T =imK_f = 3 * 1,977 * 1.86 = 11,03° C

New freezing point = 0 - 11,03 = –11,03° C

ANSWER: –11,03° C