Question

The latent heat of fusion of ice is 80 calories per gram at ${0^0}$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) ${T_f} = {0.73^0}$C
(B) ${T_f} = {1.46^0}$C
(C) ${T_f} = {1.83^0}$C
(D) None of these

Answer 1

The latent heat of fusion of ice is 80 calories per gram. This means that 80 calories of heat absorbed by one gram of ice is melting. This is the same amount of heat that was given up when the water froze. The freezing point of KCl can be found by the formula as-
${{T}_{f}^{'}} =$ ${{T_f} - \Delta {T_f}}$

Complete step by step solution:
For this type of question, first, write what is given to us and what we need to find out. Then it will be easy to get an answer.
Given :
Latent heat of fusion of ice = 80 calories per gram
Temperature = ${0^0}$C = 273.15 K
Solute = 7.45 g
Solvent (water) = 500 g
Dissociation of salt (α) = 95 %
Need to find : Freezing point of a solution of KCl in water
We know the formula for freezing point constant is -
${K_f}({H_2}O) = \dfrac{{RT_f^2{M_{solvent}}}}{{\Delta {H_{fusion}} \times m \times 1000}}$
Where ${K_f}$ = freezing point constant
${M_{solvent}}$= molar mass of solvent
${T_f}$= temperature
$\Delta {H_{fusion}}$= latent heat of fusion
By putting all the values, we get
${K_f}({H_2}O) = \dfrac{{2 \times {{(273.15)}^2} \times 18}}{{80 \times 18 \times 1000}}$
${K_f}({H_2}O) = \dfrac{{2685993.2}}{{1440000}}$
${K_f}({H_2}O) = 1.865K \cdot Kgmo{l^{ - 1}}$
Now, let's see the depression at the freezing point. The formula for it is -
$\Delta {T_f} = {K_f} \times m \times i$
Where ${K_f}$ = freezing point constant
m= molality
‘i’ = Van't Hoff factor
The molality can be found as -
$Molality = \dfrac{{Number{\text{ of moles of solute}}}}{{Weight{\text{ of solvent}}}}$
$Molality = \dfrac{{0.1}}{{0.5}}$
So, now we can find depression in freezing as-
$\Delta {T_f} = {K_f} \times m \times i$
By putting all the values
$\Delta {T_f} = 1.865 \times 0.2$
$\Delta {T_f} = {0.73^0}C$
So, now we can find freezing point as-
${{T}_{f}^{'}} =$ ${T_f} - \Delta {T_f}$
${{T}_{f}^{'}} =$ $0 - 0.73$
$\Delta {T_f} = - {0.73^0}C$
This is not in any of the options.

So, the option (D) None of these will be the right option.

Note: It must be noted that the amount of heat given out when the water freezes to ice is the same as that required to convert it into water. This heat is called the latent heat of fusion. The depression in freezing point is a colligative property. It depends upon the number of moles of solute present.