Question

# 1. Ethylene glycol [CH2(OH)CH2(OH)] is a common automobile antifreeze. Calculate the freezing point and boiling point...

1. Ethylene glycol [CH2(OH)CH2(OH)] is a common automobile antifreeze. Calculate the freezing point and boiling point of a solution containing 323 g of ethylene glycol and 1025 g of water. (Kb and Kf for water are 0.52°C/m and 1.86°C/m, respectively.)

freezing point

___  °C

boiling point

___°C

2. Calculate the molar mass of naphthalene, the organic compound in mothballs, if a solution prepared by dissolving 10.0 g of naphthalene in exactly 200 g of benzene has a freezing point 2.0°C below that of pure benzene. (Kf of benzene is 5.12°C/m.)

_____ g/mol

first calculate the molality of the solution

molality = (W/MW) (1000 / mass of solvent in g)

moality = (323 / 62.07) (1000 / 1025)

molality = 5.08 m

now Tf = Kf x m Tf = 1.86 x 5.08 Tf = 9.45

Tf = 0 - 9.45

Tf = - 9.450C Tb = Kb x m Tb = 0.52 x 5.08 Tb = 2.642

Tb = 100 + 2.642

Tb = 102.6420C

2) Tf = Kf x m Tf = Kf x W x 1000 / MW x mass of solvent in g Tf = 5.5 - 2.0 = 3.5

MW = Kf x W x 1000 / Tf x mass of solvent in g

MW = 5.12 x 10.0 x 1000 / 3.5 x 200

MW = 51200 / 700

MW = 73.14 g/mol

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