Question

An aqueous solution contains 6.09% phenylalanine (C9H11NO2) by mass. Assume the phenylalanine is non-ionic and non-volatile. Find the freezing point, boiling point, and osmotic pressure of the solution at 25 °C. Assume the solution has a density of 1.00 g/mL. The molal freezing point depression constant for water, Kfp, is -1.86 °C/m. The molal boiling point elevation constant of water, Kbp, is 0.5121 °C/m.

Answer 1

molality in moles of solute per kilogram of solventsince phenylalanine is 6.09% so in 1000 mL 60.9 g of phenylalanine is needed.

moles of phenylalanine = mass/molar mass = 60.9/165.19= 0.37 moles

so molality m = 0.37 / 1L = 0.37

Tf = Kf.m = -1.86 X 0.37 = - 0.69 K

Freezing point of water = 273K

depression in freezing point = -0.69K

therefore freezing point of the solution = 273 - 0.69 = 272.314 K

Tb = Kb*m = 0.5121 * 0.37 = 0.189

boiling point of water = 373 K

therefore boiling poing of solution = 373 + 0.189 = 373.189 K

osmotic pressure = mRT = 0.37 moles/L X 0.0821 atmLmol^-1K^-1 X (273+25)K = 9.05 atm