When 0.654 grams of a polymer were dissolved in 40.9 mL of benzene at 21.9 degrees C, the osmotic pressure was found to be 0.048 atm. Calculate the molar mass of the polymer.

Molar mass of the polymer is 8055.6 g/mol

Explanation:

Let's apply the colligative property of osmotic pressure:

π = M. R. T where

π is the osmotic pressure

M is molarity (mol / L)

R is the universal gases constant (0.082 L. atm/mol. K)

T is Absolute T° (°C + 273)

T° K = 294.9K

Let's determine our unknown value in molarity

Molarity is mol / L and mol is mass / molar mass

Molarity is (mass / molar mass) / volume, that can be written as:

(mass / molar mass). 1 / volume

M = (0.654 g / molar mass). 1 / 0.0409 L

Remember that volume must be in Litters. → 40.9 mL. 1 L/1000 mL = 0.0409 L

0.048 atm = (0.654 g / mol mass).1 / 0.0409 L. 0.082 L. atm/mol. K. 294.9K

0.048 atm / 0.082 L. atm/mol. K. 294.9K = (0.654 g / mol mass).1 / 0.0409 L

1.98*10⁻³ L/mol = (0.654 g / mol mass).1 / 0.0409 L

1.98*10⁻³ L/mol. 0.0409 L = 0.654 g / molar mass

8.12*10⁻⁵ mol = 0.654 g / molar mass

Molar mass = 0.654 g / 8.12*10⁻⁵ mol → 8055.6 g/mol