Question

Find the freezing point of a solution of 4.4 grams of naphthalene in 100 grams of benzene with Kf = 5.10oC / mol and Mr naphthalene = 128 grams / mol (freezing point of solvent = 5.5oC)​

Answer 1

Answer:

Δt = i Kf m

4.4 °C = (1) (30. °C kg mol¯1) (x / 1.50 kg)

4.4 °C = (1) (20. °C mol¯1) (x)

x = 0.22 mol

0.22 mol times 80.0896 g/mol = 17.6 g (I'll ignore sig figs and leave it at three. I'm such a rebel!)

Explanation: