1.) A protein solution containing 70.0 mg of the protein (dissolved in water) di

Question

1.) A protein solution containing 70.0 mg of the protein (dissolved in water) displays an osmotic pressure of 319.88 Pa at 310.15 K. If the volume of the protein solution is 20.0 mL, what is the molar mass of the protein?

2.) Calculate the osmotic pressure at 15°C of a solution of naphthalene (C10H8) in benzene containing 0.228 mol naphthalene per liter of solution.

3.) 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.)

Explanation / Answer

(1)

Osmotic pressure = C R T

319.88 = C * 8.314 * 1000 * 310.15

C = 1.24 * 10-4 M

Moles of protein = molarity * volume of solution i L = 1.24*10-4*20.0 / 1000 = 2.48 * 10-6 mol

Molar mass protein = mass / number of moles = 70.0 * 10-3 / 2.48 * 10-6 = 2.82 104 g. / mol

(2)

Osmatic pressure = C R T = (0.228 / 1.00) * 0.0821 * 288.15 = 5.39 atm

(3)

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

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