Question

the freezing point of water is 0.00°C at 1 atmosphere. If 14.56 grams of silver acetate, (166.9 g/mol), are dissolved in 283.7 grams of water. The molality of the solution is The freezing point of the solution is

Answer 1

We know that ΔT _f = K_f x m
Where

ΔT _f = depression in freezing point

        = freezing point of pure solvent – freezing point of solution

        = 0.00 oC - T_f

= -T_f

K _f = depression in freezing constant of water = 1.86 °C/m

m = molality of the solution

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

= ( 14.56g/(166.9(g/mol)) / 0.2837 kg

= 0.3075 m

Plug the values we get -T_f = K_f x m

= 0.57

T_f = -0.57 ^oC