Question

What is the freezing point of a solution prepared by adding 50.0 g of NaCl to 250. g of pure water? (Water has a kfp =1.86 oC/m)

Answer 1

Here,

Number of moles of NaCl = 50 / 58.5 = 0.8547

Mass of water = 250 g = 0.25 kg

Molality = number of moles / mass of solvent

              = 0.8547 / 0.25

= 3.4188 m

K_f = 1.86^oC / m

T_f = i x K_f x m (For NaCl vant-Hoff factor = 2)

         = 2 x 1.86 x 3.4188

        = 12.71 ^oC

T_f = 12.71 ^oC

To - T_f = 12.71 ^oC

0 - T_f = 12.71 ^oC

T_f = -12.71 ^oC

Freezing point = - 12.71 ^oC