Question

A solution of an enzyme is prepared by dissolving 0.138 g of the enzyme in enough solvent to make 0.250 L of solution. The resulting solution has an osmotic pressure of 1.46x10^-3 bar at 23C. Assuming that this enzyme is a non-electrolyte, compute its molecular weight.

Answer 1

P= 0.00146 bar

= (0.00146/1.01325) atm

= 0.0014 atm

T= 23.0 oC

= (23.0+273) K

= 296 K

we have below equation to be used:

P = C*R*T

0.0014 = C*0.0821*296.0

C =0.0001 M

volume , V = 0.250 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.0001*0.25

= 1.482*10^-5 mol

mass of solute = 0.138 g

we have below equation to be used:

number of mol = mass / molar mass

1.482*10^-5 mol = (0.138 g)/molar mass

molar mass = 9.31*10^3 g/mol

Answer: 9.31*10^3 g/mol