JEE MainChemistrySolutions
An aqueous solution contains 20 % by mass of a weak monoprotic acid HA (molar mass 125 g mol ⁻¹ ). If the solution freezes at 267.6 K , the percentage dissociation of the acid is ________. (Nearest integer) [Given: K_f for water = 1.8 K kg mol ⁻¹ , Freezing point of pure water = 273 K ]
Correct answer
50
Step-by-step solution
Let the mass of the solution be 100 g . Mass of solute ( HA ) = 20 g Mass of solvent (water) = 100 - 20 = 80 g = 0.08 kg Moles of solute = 20 125 = 0.16 mol Molality ( m ) = 0.16 0.08 = 2 m Depression in freezing point ( T_f ) = 273 - 267.6 = 5.4 K Using the formula T_f = i K_f m : 5.4 = i 1.8 2 i = 5.4 3.6 = 1.5 For a weak monoprotic acid, the dissociation reaction is HA H ^+ + A ^- The van't Hoff factor i = 1 + , where is the degree of dissociation. 1.5 = 1 + = 0.5 Percentage dissociation = 0.5 100 = 50 % Answer: