Question

The normal freezing point of a certain liquid X is 6.60 °C, but when 27. g of potassium bromide (KBr) are dissolved in 300. g of X the solution freezes at 4.9 °C instead. Use this information to calculate the molal freezing point depression constant K, of X. Be sure your answer is rounded to the correct number of significiant digits. Ķ, = 0 °C kg x s ?

Answer 1

1)

mass of KBr dissolved = 27. g

moles of KBr = 27.7 / 119 = 0.227 mol

molality = moles / mass of solvent in kg

             = 0.227 / 0.300

molality = 0.756 m

delta Tf = i x Kf x m

6.60 - 4.90 = 2 x Kf x 0.756

Kf = 1.1 ^oC kg/mol