Question

Calculate the depression in the freezing point of water when 10 g of \(CH_3CH_2CHClCOOH\) is added to 250 g of water. \(K_a = 1.4 × 10^{–3} , K_f = 1.86 K kg mol^{–1} .\)

Answer 1

The correct answer is: 0.65 K
Molar mass of \(CH_3CH_2CHClCOOH=15+14+13+35.5+12+16+16+1\)
\(=122.5gmol^{-1}\)
∴No. of moles present in 10 g of \(CH_3CH_2CHClCOOH=\frac{10g}{122.5gmol^{-1}}\)
=0.0816mol
It is given that 10 g of \(CH_3CH_2CHClCOOH\) is added to 250 g of water. 
∴Molality of the solution,\(=\frac{0.0186}{250} \times 1000\)
\(=0.3264molkg^{-1}\)
Let α be the degree of dissociation of \(CH_3CH_2CHClCOOH\)
\(CH_3CH_2CHClCOOH\) undergoes dissociation according to the following equation:
\(CH_3CH_2CHClCOOH↔CH_3CH_2CHClCOO^-+H^+\)
Initial conc.      \( CmolL^{-1}\)           0       0
At equilibrium    \(C(1-α)\)       \(Cα\)     \(Cα\)
\(∴K_a=\frac{Cα.Cα}{C(1-α)}\)
\(=\frac{Cα^2}{1-α}\)
Since \(α\) is very small with respect to \(1, 1 - α ≈1\)
Now, \(K_a=\frac{Cα^2}{1}\)
\(⇒K_a=Cα^2\)
\(⇒\sqrt{\frac{K_a}{C}}\)
\(=\sqrt{\frac{1.4 \times 10-3}{0.3264}} \,\,\,\,(∴K_a=1.4 \times 10^{-3})\)
=0.0655
Again,
\(CH_3CH_2CHClCOOH↔CH_3CH_2CHClCOO^-+H^+\)
Initial conc.               1            0       0
At equilibrium    \((1-α)\)       \(α\)     \(α\)
Total moles of equilibrium \(= 1 - α+ α+ α\)
\(= 1 + α\)
\(∴i=\frac{1+α}{1}\)
\(=1+α\)
=1+0.0655
=1.0655
Hence, the depression in the freezing point of water is given as:
\(ΔT_f=i.K_fm\)
\(=1.0655 \times 1.86K \,Kg mol^{-1} \times 0.3264 mol Kg^{-1}\)
=0.65 K