Question

(a) Calculate the freezing point depression and osmotic pressure at 25⁰C of an aqueous solution containing 1.0g/L of a protein (molar mass 9 x 10⁴ g/mol) if the density of the solution is 1g/ cm³.(b) Considering your answer to part a which colligative property freezing point depression or osmotic pressure would be better used to determine the molar masses of large molecules. Explain.

Short answer

(a) Osmotic pressure of the solution is 2.036.8 x 10^-4torr

The freezing point of the solution is 0.2046 x 10^-40C

(b)It is because even in dilute solution the osmotic pressure values are high and can be measured accurately and it can be measured at room temperature.

Step-by-step solution

Step 1- Osmotic pressure

Molarity of solution = Moles of solute/ litre of solution

= 1.0 g/l / 9 x 10⁴ g/mol

= 0.11 x 10^-4 mol/ L

Temperature in kelvin = 25 +273= 298K

Osmotic pressure= M R T

= 0.11 x 10^-4 x 0.08206 x 298

= 2.68 x 10^-4atm

Or in Torr = 2.68 x 10 ^-4 x 760 = 2036.8 x 10^-4Torr

Step 2- Freezing point of the solution

$∆$T_f = K_f x m

K_ffor water is 1.86⁰C/m

m= molality

Mass of 1L (1000cm³) of this solution = density x volume of solution

= 1000g

Molarity = 0.11 x 10^-4 mol of protein in 1L of solution

Mass of protein = molarity x molar mass

= 0.11 x 10^-4 x 9 x 10⁴=0.99 g

1000g of solution – 0.99 g of protein = 999.01 g of solvent

Molality= 0.11 x 10^-4/ 0.999 kg

= 0.110 x 10^-4m

Now depression in freezing point = molality x K_f

= 0.110 x 10^-4x 1.86

= 0.2046 x 10^{-4 0}C

Step 3- Osmotic pressure

Osmotic pressure would be used to determine molar masses of large macromolecules.

It is because even in dilute solution the osmotic pressure values are high and can be measured accurately and it can be measured at room temperature.