Question

Osmotic pressure of a solution containing 2 g of a protein (soluble in water) per 300 cm³ of the solution is 20 mm of Hg at 27°C. Calculate the molecular mass of protein.

(R = 0.0821 L atm K⁻¹ mol⁻¹)

Answer 1

Given:

Mass of protein (solute) = 2 g

Volume of solution = 300 cm³ = 0.300 L

Osmotic pressure π = 20 mm Hg = 20760 ≈ 0.02632 atm

Temperature T = 27°C = 300 KT

Gas constant R = 0.0821 L atm K⁻¹ mol⁻¹

By using osmotic pressure formula

`pi = n/V RT`

⇒ `n = (pi V)/(RT)`

`n = (0.02632 xx 0.300)/(0.0821 xx 300)`

= `0.007896/24.63`

= 3.206 × 10⁻⁴ mol

Molar mass = `"Mass of solute"/"Moles"`

= `2/(3.206 xx 10^-4)`

= 6240 g/mol

∴ The molecular mass of the protein is 6240 g/mol