A protein sol containing $8.0 g L^{- 1}$ of the protein supports a water column of height $3.2 cm$ at osmotic equilibrium at $300 K$ . Take the density of water as $1000 kg m^{- 3}$ , $g = 9.8 m s^{- 2}$ and $R = 8.314 J mo l^{- 1} K^{- 1}$ . Find the molar mass of the protein in $g mo l^{- 1}$ , to the nearest integer.

Numerical answer type, answer in g/mol, tolerance 5.

Answer: 63628 g/mol

Osmotic pressure =

ρ g h = 1000 \times 9.8 \times 0.032 = 313.6

Pa. With

π = (w / M) RT

and

w = 8.0 kg m^{- 3}

M = wRT / π = 8.0 \times 8.314 \times 300/313.6 = 63.63 kg mo l^{- 1}

. This is 63628 g/mol.