What is the percentage dissociation of 0.8ml of Acetic acid (density is 1.04g/ml) which is dissolved in 1.2L of water if the observed Depression in freezing point is 0.0228 K? ( K_f for water = 1.86Kkg/mol.)

Step-by-step solution

Mass of acetic acid = Volume Density = 0.8 1.04 = 0.832 g Molar mass of acetic acid ( CH ₃ COOH ) = 60 g/mol Number of moles of acetic acid = 0.832 60 = 0.01387 mol Mass of water = 1.2 L 1 kg/L = 1.2 kg Molality ( m ) = Moles of solute Mass of solvent in kg = 0.01387 1.2 = 0.01156 mol/kg Theoretical depression in freezing point ( T_ f( th ) ) = K_f m = 1.86 0.01156 = 0.0215 K Observed depression in freezing point ( T_f ) = 0.0228 K van't Hoff factor ( i ) = Observed T_f Theoretical T_ f( th ) = 0.0228 0.0215 1.06 F