Question

A  0.2 m aqueous solution of KCl freezes at -0.68°C. Calculate ‘i’ and the osmotic pressure at 0°C. Assume the volume of solution to be that of pure H₂O and  K_f for H₂O is 1.86 K Kg/mol.

Answer 1

(ΔT_f)normal = K_f x m = 1.86 x 0.2= 0.372

i= Observed colligative property/Normal colligative property

= 0.68/0.372

Observed osmotic pressure= i x Normal osmotic pressure

= i x c RT

= 1.83 x 0.2 x 0.082 x 273

= 8.2 atm