What is the increase of the boiling point of a solution of 2.0 mol CaCl2 in 1 kg of water if the change is 0.51 c/mol of particles?
Expert Answer
Apply Colligative properties
We can favourincrease in BP via:
dTb = Kb*m*i
where i = total ions in solution, in this case CaCl2 = Ca+2 + 2Cl- = 1+2 = 3 ions
m = molalty = mol of solute / kg of solvent
1 kg of solvent, 2 mol
so
molality = 2/1 = 2
now;
dTb = 0.51*2*3 = 3.06 ºC
this is the increase in boiling point, i..e it will boil at 103.06 C