The osmotic pressure of 1.00 g of a polymer dissolved in benzene to give 200 mL of solution is 1.50 kPa at 25 $\circ$ C.Estimate the average molar mass of the polymer.The gas law constant is 0.0821 L.atm.mol^-1.K^-1.

Question

Quizplus · Accepted Answer

We can use the formula:

Π = MRTi

where Π is the osmotic pressure, M is the molar mass of the polymer, R is the gas law constant, T is the temperature in Kelvin, and i is the van't Hoff factor (usually assumed to be 1 for organic compounds).

We first need to convert 1.50 kPa to atm:

1.50 kPa = 0.0147 atm

We also need to convert 25°C to Kelvin:

25°C = 298 K

Plugging in the given values:

0.0147 atm = M (0.0821 L·atm/(mol·K)) (298 K)

Solving for M:

M = 8260 g/mol

Therefore, the average molar mass of the polymer is 8260 g/mol, which corresponds to choice B.

The Osmotic Pressure of 1 ∘\circ∘ CEstimate the Average Molar Mass of the Polymer

The Osmotic Pressure of 1 $\circ$ CEstimate the Average Molar Mass of the Polymer