Question

The osmotic pressure of a 6.613 g dm-3 solution of polystyrene in toluene was measured at 25oC, and the pressure, expressed in terms of the height of the solvent, was 1.910 cm. The density of toluene at this temperature is 1.004 g cm3. Calculate the molar mass of the polymer Using PI=[B]RT

Where [B]= nb/V as the molar concentration of the solution.

Answer 1

Apply

Osmotic pressure equation

PI = M*R*T

M = molarity

so

C = 6.613 g/dm3 = 6.613 g/L

T = 25°C = 298K

toluene is the solvent

The pressure exerted:

P = rho*g*h = (1004 kg/m3)(9.8 m/s2)(1.91*10^-2) = 187.92872 Pa

P = 187.93 Pa / 101325 Pa / Atm = 0.0018547 atm

PI = M*R*T

0.0018547 = M * 0.082*298

M = 0.0018547/( 0.082*2989)= 0.00000756717 M

assume a basis of 1 LITER of solution

so

6.613 g of polystyrene is present

mol = 0.00000756717*1 = 0.00000756717 mol of polystyrene

so

MW = 6.613 g /0.00000756717 mol

MW = 873906.625595 g/mol