1. The boiling point constant and freezing point depression constant of water ar
ID: 568773 • Letter: 1
Question
1. The boiling point constant and freezing point depression constant of water are given below: Kt (K mola') 1.85 KA (K molar) 0.51 T, (k) 1 373 K or 100 C Water | 273 Kor 0 C If a chemical is added into solid ice at 0°C, the ice will melt or turn into water due to the lowering of the freezing point of water. One way to remove ice on roadway is to spray salts onto the surface. In the laboratory, one can make a -20 °C ice-water bath by adding some salt (e.g. CaCl2) into a bucket of water-ice. If 0.5 kg of CaCl2 has been dissolved and the liquid portion of the bucket is 1 kg (CaCl2 will dissociate into Ca2 and 2 Cl in water), a. What is the molal concentration of the salt solution? b. What is the freezing point of the water solution or the temperature of ice-water solution?Explanation / Answer
a) molality of salt = moles of salt / mass of solvent in kg
= [500g /111g/mol] / 1 kg
=4.5045 mol/kg
b) First we calculate the depression in freezing point , delta Tf
Delta Tf = i x Kf x m
where i = van't Hoff factor
For CaCl2 i = 3 ( dissociated into 3 ions)
Thus delta Tf = 3 x 1.85x4.5045
=25C
Thus the freezing point of solution = normal freezing point - delta Tf
= 0 C - 25 C
= -25C
or = 273 -25 K
= 248 K
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