Question

The following data is available for a solution of lysozyme.

Lysozyme (g/L)	Osmotic Pressure (mmHg)
4.06	4.4
6.13	6.5
8.18	8.65
10.42	10.75
12.26	12.55
14.26	14.55
16.31	16.3
18.38	18.35
20.44	20.15

Determine the molar mass and osmotic virial coefficient for lysozyme.

Answer 1

osmotic pressure is a colligative property ,

osmotic pressure =c*R*T

where c=concentration of lysozyme=mass in g/L/M ,M=molar mass of lysozyme in g/mol

R=universal gas constant =0.0821 L atm/K.mol

T=temperature=25 deg C=273+25=298K

So =(4.06 g/L/M)*(0.0821 L atm/K.mol)*298K

or,=99.331 atm/M

=4.4 mmhg=4.4/760 =0.00579 atm

0.00579 atm=99.331 atm/M

or,M=17155.613 g/mol(answer)

For virial coefficient,osmotic pressure =c*R*T(1/M+B c) B=virial coefficient

putting values from the data table,

0.00579 atm=(4.06 g/L)*(0.0821 L atm/K.mol)*298K(1/M +B*4.06g/L)

or,0.00579 atm=99.331 g atm/mol(1/M +B*4.06g/L)

(1/M +B*4.06g/L)=5.82*10^-5 mol/g...................(1)

second data gives

6.5/760 atm=(6.13 g/L)*(0.0821 L atm/K.mol)*298K(1/M +B*6.13g/L)

0.00855 atm=149.975 g atm/mol (1/M +B*6.13g/L)

5.7*10^-5 mol/g= (1/M +B*6.13g/L)...............(2)

eqn (1)-eqn(2) gives,

0.12*10^-5 mol/g=B(4.06 g/L-6.13g/L)

or,0.12*10^-5 mol/g=B(-2.07 g/L)

B=0.058*10^-5 mol L/g^2

from eqn (1), putting B value M can be obtained

(1/M +B*4.06g/L)=5.82*10^-5 mol/g.

1/M+(0.058*10^-5 mol L/g^2)*(4.06g/L)=5.82*10^-5 mol/g.

1/M+0.235*10^-5 mol/g=5.82*10^-5 mol/g.

1/M=5.58*10^-5 mol/g

M=0.179*10^5 g/mol(answer)=17921.147 g/mol