question_answer

  0.6 mL of acetic acid () having density 1.06 g  is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the van't Hoff factor and the dissociation constant of acid.                                                                                           [3]

Answer:

  Mass of solute volume x density = 0.6 x 1.06 = 0.636 g Mass of solvent (water)  = volume x density = 1000 x 1    = 1000                              (molar mass of solute) = 60 g mol 1 (depression in freezing point) = 0.0205° (molal depression constant of water) i = van't Hoff factor = [1 + ( y ? 1 ) x ] = ( 1 + x )                         y = total number of ions from 1 unit of solute x = degree of ionisation = 1.04 Thus,   i = (1 + x) = 1.04 x = 0.04                                                            [1] By Ostwald's dilution law for weak acid where, C = molar concentration of ()                                                                                                                                                    [1]