Question

0.15g of purified protein os dissolved in water to give 2.0ml of solution. The osmotic pressure is found to be 8.6torr at 25 degrees Celsius. Calculate the molecular mass of the protein

Answer 1

P = C*R*T, where P = 8.6 torr = 1146,57 Pa - the osmotic pressure R = 8,314 J*K^-1*mol^-1 - universal gas constant, T = 25 + 273,15 = 298,15 K - temperature, C - molar concentration. C = m/(M*V), where m - mass of protein(0,15*10^-3 kg), M - the molecular mass of the protein, V - volume of solution(2*10^-6 m³)

P = m*R*T/(M*V) and M = m*R*T/(P*V) = 162,145 kg/mol = 162,145 kDa