An aqueous solution of 0.0100 g of a protein in 10.0 mL of aqueous solution at 2

Question

An aqueous solution of 0.0100 g of a protein in 10.0 mL of aqueous solution at 20 degree C shows a rise of 5.22 cm in the apparatus shown in the right figure. The density of the solution is 0.998 g cm^-3. You may find other needed information on textbook table 10.8 (for 5/e, table 9.8). (a) What is the molar mass of the protein? (b) What is the normal freezing point of the solution? (c) What is the normal boiling point of the solution? (d) Which colligative property is best for measuring the molar mass of these large molecules? Please reason for your answer.

Explanation / Answer

(a) pressure change = iMRT

mass of solution = 10 ml

M = molarity = mass of solute/molar mass of protein x volume of solution

So,

molar mass of protein = 1 x 0.01 x 0.08205 (273 + 20)/(5.22 x 10/760) x 0.01 = 350.02 g/mol

(b) mass of solution = 10 x 0.998 = 9.98 g = 0.00998 kg

dTf = iKfm (m = molality)

0 - dTf = solution freexing point

normal freezing point of solution = 0 - (1 x 1.86 x 0.01/350.02 x 0.00998 kg) = -0.005 oC

(c) 100 + dTb = solution boiling point

dTb = iKbm (m = molality)

normal boiling point of solution = 100 + (1 x 0.51 x 0.01/350.02 x 0.00998) = 100.0015 oC

(d) Depression in freezing point is best way to detemine molar mass of large molecule as the smallest of depression is shown clearly in the solution accurately.

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