1.) A solution of potassium chloride (KCl) is prepared by dissolving 9.63 g of t

Question

1.) A solution of potassium chloride (KCl) is prepared by dissolving 9.63 g of the salt in 1.00 kg water (H2O). What is the freezing point of this solution? (The freezing point of pure water is 0°C (exactly) at 1 atm pressure; the molal freezing-point-depression constant for water is 1.86°C/m.)

2.) A solution of sodium bromide (NaBr) is prepared by dissolving 6.46 g of the salt in 1.00 kg water (H2O). What is the freezing point of this solution? (The freezing point of pure water is 0°C (exactly) at 1 atm pressure; the molal freezing-point-depression constant for water is 1.86°C/m.)

3.) What is the boiling point (in °C) of a solution of 9.47 g of I2 in 63.3 g of toluene, assuming the I2 is nonvolatile? (For toluene, Tb = 110.63°C and Kb = 3.40°C·kg/mol.)

Explanation / Answer

(a)

Moles of KCl, n = mass / molar mass = 9.63 / 74.5 = 0.129 mol

Molality, m = moles of solute / mass of solvent in kg = 0.129 / 1.00 = 0.129 m

Kf of water = 1.86 0C/m

Freezing point of water, T0 = 00C

Freezing point of solution, Tf = ?

We know that,

T0 - Tf = Kf * m

0 - Tf = 1.86 * 0.129

Tf = Freezing point of solution = - 0.240 0C

(2)

Moles of NaBr, n = mass / molar mass = 6.46 / 103 = 0.0627 mol

Molality, m = moles of solute / mass of solvent in kg = 0.0627 / 1.00 = 0.0627 m

We know the formula,

0 - Tf = 1.86 * 0.0627

Tf = - 0.117 0C

(3)

Moles of I2 = 9.47 / 254 = 0.0373 mol

Mass of solvent = 63.3 g. = 0.0633 kg.

Molality = moles of solute / mass of solvent in kg = 0.0373 / 0.0633 = 0.589 m

Tb = 110.63 0C and Kb = 3.400C.kg/mol

We know the formula,

T - Tb = Kb * m

T - 110.63 = 3.40 * 0.589

T = Boiling point of solution = 112.63 0C

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