Question

An aqueous solution contains 0.63 g of protein in 300 cm³ of water. The osmotic pressure of the solution at 300 K is 1.29 × 10⁻³ atm. Calculate the molecular mass of protein.

(Given R = 0.0821 Lit atm K⁻¹ mol⁻¹)

Answer 1

Given: w = 0.63 g

V = 300 cm³ or 0.3 litre

R = 0.0821 Lit atm K⁻¹ mol⁻¹

T = 300

π = 1.29 × 10⁻³ atm

To find: m = ?

Formula: `πV = w/m RT`

`1.29 xx 10^-3 xx 0.3 = 0.63/m xx 0.0821 xx 300`

m = `(0.63 xx 0.0821 xx 300)/(1.29 xx 10^-3 xx 0.3)`

m = `(15.5169)/(0.387 xx 10^-3)`

m = `(15.5169)/(0.000387)`

m = 40,095.35 g mol⁻¹