Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10–3, Kf = 1.86

Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water. Ka = 1.4 × 10^–3, K_f = 1.86

K kg mol^–1.

Solution:

Ans. Molar mass of CH3CH2CHClCOOH

=12 + 3 + 12 +2+12 + 1 35.5 +12 +16+16+1  

= 122.5 g m

Number of moles in 10 g of CH3CH2CHCICOOH = 10/122.5 mole

= 8.16 × 10^-2 mole

Density of water = 1 g/ ml

250 g of water  = 250ml

To convert in liter divide by 1000 we get

250 ml = 250/1000 Liter   = 0.25 Liter  

                     = 10/122.5 mole

                     = 8.16 × 10^-2 mole

Plug the values we get

Molarity = 8.16 × 10 ^-2 / 0.25   = 0.3264 M

Let α is the degree of dissociation

Mass of solvent = 250g = 0.25 Kg

 

Plug the values we get

Molality = 8.16 × 10 ^-2 / 0.25   = 0.3264 m