Question

the freezing point of 1.0 Liters of water was determined to be -3.5°C. how many grams of NaCk have been added to the water? (Assume that the density of water is 1.0 g/mL)

Answer 1

Solution :-

volume of water = 1.0 L *1000 ml / 1 L = 1000 ml

mass of water = 1000 ml * 1 g/ml = 1000 g

change in freezing point Delta Tf = 3.5 C

Delta Tf = Kf* m * i

Kf= 1.86 C/m

i = 2 Vant Hoff Factor

Lets put the values in the formula

3.5 C = 1.86 C/m * m * 2

3.5 C / 1.86 C per m * 2 = m

0.9408 m = m

so the molality of the solution is 0.9408 m

now using the molality and mass of solvent lets calculate the moles of NaCl

moles = molality * kg solvent

         = 0.9408 mol per kg * 1 kg

         =039408 mol NaCl

now lets calculate the mass of NaCl

mass= mole * molar mass

        = 0.9408 mol * 58.443 g per mol

        = 55.0 g NaCl

So the amount of NaCl added in the water is 55.0 g