Question

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.

Answer 1

Molar mass of the polymer is 8055.6 g/mol

Step-by-step 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