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Question:
Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water. K_a = 1.4 x 10^-3, K_f = 1.86 K kg mol^-1.

Answer:

Molar mass of CH₃CH₂CHClCOOH

15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1

= 122.5 g/mol

∴ Moles of CH₃CH₂CHClCOOH = 10g / 122.5 g/mol

= 0.0816 mol

Therefore molality of the solution

= (0.0816 x 1000) / 250

= 0.3265 mol kg^-1

Now if a is the degree of dissociation of CH₃CH₂CHClCOOH,

So, K_a = (Cα x Cα) / (C (1-α))

K_a= Cα²/ (1-α)

Since α is very small with respect to 1, 1 - α = 1

K_a= Cα²

α = √K_α / C

Putting the values ,We get

α = √1.4 x 10^-3/ 0.3265

= 0.0655

Now at equilibrium,the van’t hoff factor i = 1-α +α +α/1

= 1 + 0.0655

= 1.0655

Hence, the depression in the freezing point of water is given as:

Therefore ΔT_f = i K_f m v

= 1.065 v x 1.86 x 0.3265

= 0.647°

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Question:
Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water. K_a = 1.4 x 10^-3, K_f = 1.86 K kg mol^-1.

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Question: Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 x 10-3, Kf = 1.86 K kg mol-1.

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Question:
Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water. K_a = 1.4 x 10^-3, K_f = 1.86 K kg mol^-1.