Question

At 25 ^oC a 0.50 g sample of polyisobutylene (a polymer used in synthetic rubber) in 100.0 mL of benzene solution has an osmotic pressure that supports a 5.1 mm column of solution (p = 0.88 g/mL). What is the molar mass of the polyisobutylene? (For Hg, p = 13.6 g/mL)

Answer 1

At 25 oC a 0.50 g sample of polyisobutylene (a polymer used in synthetic rubber) in 100.0 mL of benzene solution has an osmotic pressure that supports a 5.1 mm column of solution (p = 0.88 g/mL). What is the molar mass of the polyisobutylene? (For Hg, p = 13.6 g/mL)

Two laws governing the osmotic pressure of a dilute solution were discovered by the German botanist W. F. P. Pfeffer and the Dutch chemist J. H. van’t Hoff:

1. The osmotic pressure of a dilute solution at constant temperature is directly proportional to its concentration.

2. The osmotic pressure of a solution is directly proportional to its absolute temperature.

These are analogous to Boyle's law and Charles's Law for gases. Similarly, the combined ideal gas law,

PV = nRT, has an analog for ideal solutions:

πV = nRTi

where: π = osmotic pressure; V is the volume; n is the number of moles of solute;

R =0 .08206 L atm mol-1 K-1 OR R = 62.363 L mmHg K−1 mol−1 the molar gas constant;

T is absolute temperature;

i = Van 't Hoff factor, which equals 1 for non-ionics

where n is number of moles,

so we first need to calculate π in mmHg:

(5.1 mm solution) (0.88 g/mL solution) / (13.6 g/mL Hg) = 0.33 mmHg.

Now calculate the number of moles:

n = πV/RT = (0.33 mmHg) (100mL) / (273.15 + 25 K) (62.3637 L*mmHg/K*mole) (1000 mL/L);

n= 33/ 18584382.6

n = 1.775 x 10^-6 moles

Since there was 0.50 g of the polymer in solution…

(0.50 g) / (1.775 x 10^-6 moles) = 2,81,690.14 g / mole

Molar mass of polyisobutylene is 2,81,690.14 g / mole