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