Question

1. the freezing point depression constant of carbon tetrachloride is 30 degrees c/m. the density of carbontetrachloride is 1.584g/cm^3. The normal freezing point of carbon tetrachloride is -23 degrees C.

a. calculate the freezing point of a solution made by mixing 12.28g of solid iodine with 250 ml of carbon tetrachloride. (show all work)

b.calculate the freezing point of a solution made by mixing 12.28g of solid naphthalene (C10H8) with 250 ml of carbon tetrachloride. (show all work)

Answer 1

Freezing point depression

Tf - Ts = i x Kf x m

Freezing point of CCl4 - freezing point of solution = van't Hoff factor x freezing point depression constant x Molality

Part a

i = 1 (I2 doesn't break into ions in CCl4 solution)

Kf = 30 °c/m

Mass of CCl4 = volume x density

= 250 mL x 1.584 g/cm3 x 1cm3/1 mL

= 396 g x 1kg/1000g = 0.396 kg

m = moles of I2/mass of CCl4

= mass of I2/(molecular weight of I2 x mass of CCl4)

= 12.28g /( 253.81 g/mol x 0.396 kg)

= 0.122 mol/kg

-23 - Ts = 1 x 30 x 0.122

-Ts = 3.66 + 23

Ts = - 26.66 °C

Part b

i = 1 (naphthalene doesn't break into ions in CCl4 solution)

Kf = 30 °c/m

Mass of CCl4 = volume x density

= 250 mL x 1.584 g/cm3 x 1cm3/1 mL

= 396 g x 1kg/1000g = 0.396 kg

m = moles of naphthalene /mass of CCl4

= mass of naphthalene /(molecular weight of naphthalene x mass of CCl4)

= 12.28g /( 128 g/mol x 0.396 kg)

= 0.2422 mol/kg

-23 - Ts = 1 x 30 x 0.2422

-Ts = 7.266 + 23

Ts = - 30.266 °C