Question

Calculate the freezing point of a solution containing 30 grams of KCL and 3300.0 grams of water. The molal-freezing-point-depression constant (Kf) for water is 1.86 degrees Celsius/M

Answer 1

We know that ΔT _f = iK_f x m
Where

ΔT _f = depression in freezing point

        = freezing point of pure solvent – freezing point of solution

        = 0 - T_f

=  - T_f

K _f = depression in freezing constant = 1.86 oC/m

i= vanthoff’s factor = 2 (KCl ---> K⁺ + Cl^- )

m = molality of the solution

    = ( mass / Molar mass ) / weight of the solvent in Kg

= (30 g / 74.5(g/mol)) / 3.300 kg

= 0.122 m

Plug the values we get  - T_f = 2x1.86x0.122 = 0.454

T_f = - 0.454 oC