Question

What are the freezing point and boiling point of a solution prepared by dissolving 18.6g of CaCl₂ in 200.0g of water? (k_f = -1.86�/m and k_b= 0.512�/m)?

Answer 1

Let us calculate moles of CaCl2

Molar mass of CaCl2 = 110.98 g/mol

moles of CaCl2 = 18.6 g CaCl2 x( 1 mole CaCl2/ 110.98g CaCl2)

                         = 0.168 moles

Then convert grams of water to kilograms of H2O

200g x ( 1kg / 10^3 g) = 0.200 kg

Calculate molality

molality = moles of solute/ kg of solvent

             = 0.168/0.200

             = 0.84 mol/kg

For boiling point we will use the relation,

dela Tb = i(Kb)(m) [taking i = 3 (Van't Hoff factor))

           = 3(0.512 C/m)(0.84 m)

           = 1.29 C

Boiling point of solution = 1.29 + 100.0 C (boiling point of water is 100 C which is to be added)

                                      = 101.29 C

Boiling point of solution is 101.29 C

Freezing point of solution calculation, we will use the relation,

dela Tf = i(Kf)(m) {taking i = 3) we have,

           = 3(-1.86C/m)(0.84m)

          = -4.67 C

Freezing point of solution is -4.67 C