Question

Calculate the dipression in the freezing point of water when 10g of CH₃CH₂CHClCOOH is added to 250g of water. Ka = 1.4 × 10⁻³, Kf = 1.86 K kg/mol

Answer 1

The depression in the freezing point of water when 10g of CH3CH2CHClCOOH is added to 250g of water is 0.607736 K.

Step-by-step explanation:

To calculate the depression in the freezing point of water when CH3CH2CHClCOOH is added, we need to determine the molality of the solution first, and then apply the freezing point depression formula.

First, calculate the moles of CH3CH2CHClCOOH (molar mass = 122.5 g/mol):

• moles = 10g / 122.5 g/mol = 0.08163 mol

Then calculate the molality (m), knowing that molality is moles of solute per kg of solvent:

• molality (m) = moles / 0.250 kg = 0.08163 mol / 0.250 kg = 0.32652 m

Because CH3CH2CHClCOOH is a weak acid and partially dissociates, we take into account its dissociation constant (Ka). Assuming simplified complete dissociation for this scenario:

• i = 1 (since it gives only one type of ion in solution)

Finally, apply the formula for freezing point depression (ΔTf):

ΔTf = kf * m * i

• ΔTf = 1.86 K kg/mol * 0.32652 m * 1 = 0.607736 K

Therefore, the depression in the freezing point of water is 0.607736 K.