Question

An aqueous solution of a non-volatile and non-electrolytic substance boils at 100.5°C. Calculate the osmotic pressure of this solution at 27°C. K_b for water per 1000 g = 0.50.

Answer 1

For the solution of the given substance,

ΔT_b = 100.5 − 100

ΔT_b = 0.5°

∵ ΔT_b = K_bm

∵ `m = (Delta T_b)/K_b`

= `0.5/0.50`

= 1

Thus, the given solution is a molal solution, i.e., 1 mole of the solute is present in 1000 g of the solvent. Since the solvent is water, its density may be taken to be equal to one.

∴ Volume of solution = Volume of solvent

= `1000/1`

= 1000 mL

= 1 L

Thus, n = 1, V = 1 L, T = 27°C = 300 K and R = 0.0821 L atm K⁻¹ mol⁻¹ 

∵ πV = n RT

∴ `pi = (n R T)/V`

= `(1 xx 0.0821 xx 300)/1`

= 24.63 atm

Hence, the osmotic pressure of the given solution is 24.63 atm.