Question

The latent heat of fusion of ice is 80 calories per gram at${0^o}C$. What is the freezing point of a solution of KCl in water containing 7.45 grams of solute in 500 grams of water, assuming that the salt is dissociated to the extent of 95%?
A. ${T_f} = {0.73^o}C$
B. ${T_f} = {1.46^o}C$
C. ${T_f} = {1.83^o}C$
D. None of these

Answer 1

the freezing point of a substance is defined as the dynamic equilibrium which is present in the solid phase with the liquid phase. It is at this temperature that the vapor pressure present in the liquid phase gets equal to the vapor pressure present in the solid phase. Now if we add some amount of solute in it then we will observe the decrease in the vapor pressure due to which the freezing point also decreases.

Complete step by step answer:
According to the Raoult law we see when the nonvolatile solute gets added to the pure solvent the freezing point of the solvent decreases.
Let the freezing point of the pure solvent be${T^o}_f$ and the freezing point of solution when we add the non-volatile solute is be${T_f}$
The formula for depression in freezing point is:
$\Delta T = {T^o}_f - {T_f}$
We know that this property is proportional to the molality of solution. So,
$\Delta T \propto m$
$\Delta T = {K_f}m$
Here${K_f}$ is the depression in freezing point or cryoscopic constant.
This constant is depending upon the nature of solvent so we can write the above equation as:$= 1.865K.Kg/mol{0^o}C - 1.865 \times 1.95\left( {\dfrac{{7.45}}{{\left( {\dfrac{{74.5}}{{0.5}}} \right)}}} \right)$
$= - 0.73{K_f}^oC{T^o}_f$
Latent heat of fusion= 80 cal per gram
R= 2 cal per K mol
Substituting the values in the formula we get:
${K_f}({H_2}O) = \dfrac{{2 \times {{(273.15)}^2} \times 18}}{{80 \times 18 \times 1000}} = 1.865K.Kgmo{l^{ - 1}}$
Depression in freezing point is :
$\Delta T = {T^o}_f - {T_f}$
$\Delta T = {T^o}_f - {T_f} = i{K_f}m$
${T^o}_f = {0^o}C$
$i = 1.95$
${K_f} = 1.86$
$m = \left( {\dfrac{{7.45}}{{\left( {\dfrac{{74.5}}{{0.5}}} \right)}}} \right)$
Substituting the values in the formula we get,
$\Delta T = {T^o}_f - {T_f} = i{K_f}m = {0^o}C - 1.865 \times 1.95\left( {\dfrac{{7.45}}{{\left( {\dfrac{{74.5}}{{0.5}}} \right)}}} \right) = - {0.73^o}C$

So the correct option is option D

Note: the vapor pressure of the solution will decrease on adding volatile solute in the volatile solvent. The properties which depend on the decrease in vapor pressure are relative lowering of vapor pressure of solvent, the depression of freezing point of solvent, osmotic pressure and elevation of boiling point of solvent. These properties are the colligative properties which help in determining the molar mass of solute.