Question

Assuming $ 100\% $ ionization, the increasing order of the freezing point of the solution will be:
A) $ 0.10mol/kgB{a_3}{(P{O_4})_2} < 0.10mol/kgN{a_2}S{O_4} < 0.10mol/kgKCl $
B) $ 0.10mol/kgKCl < 0.10mol/kgN{a_2}S{O_4} < 0.10mol/kgB{a_3}{(P{O_4})_2} $
C) $ 0.10mol/kgN{a_2}S{O_4} < 0.10mol/kgB{a_3}{(P{O_4})_2} < 0.10mol/kgKCl $
D) $ 0.10mol/kgKCl < 0.10mol/kgB{a_3}{(P{O_4})_2} < 0.10mol/kgN{a_2}S{O_4} $

Answer 1

Hint: In order to arrange them in increasing order of their freezing point, we must calculate the Van’t Hoff factor. As here $ 100\% $ ionization is taking place, we have to calculate its degree of dissociation which is further used to calculate the freezing point.
Formula used
 $ {T_f} = 273.15 - \Delta {T_f} $
And,
 $ \Delta {T_f} = i \times {K_f} \times M $
Where, $ i $ = degree of dissociation
 $ {K_f} $ = molal freezing point depression constant
 $ M $ = concentration of solution
 $ {T_f} $ = freezing point of the solution.

Complete step by step solution
Freezing point of a solution is defined as the temperature at which that solution gets converted into a solid. According to the question, there is $ 100\% $ ionization, so firstly we have to calculate the degree of dissociation of every solution.
In case of $ B{a_3}{(P{O_4})_2} $ ,
Degree of dissociation ( $ i $ ) = $ 5 $
So, the depression in freezing point will be given as:
 $ \Delta {T_f} = i \times {K_f} \times M $
Where, $ i $ = degree of dissociation
 $ {K_f} $ = molal freezing point depression constant
And $ M $ = concentration of solution
So, depression in freezing point in $ B{a_3}{(P{O_4})_2} $ will be:
 $ \begin{gathered}
   \Rightarrow \Delta {T_f} = 5 \times 1.86 \times 0.1 \\
   \Rightarrow 0.93 \\
\end{gathered} $
So, freezing point can be calculated by formula:
 $ {T_f} = 273.15 - \Delta {T_f} $
So, freezing point of $ B{a_3}{(P{O_4})_2} $ will be:
 $ \begin{gathered}
  {T_{f(1)}} = (273.15 - 0.93)K \\
   \Rightarrow 272.22K \\
\end{gathered} $
Now, in the case of $ N{a_2}S{O_4} $ ,
Degree of dissociation ( $ i $ ) = $ 3 $
So, depression in freezing point in $ N{a_2}S{O_4} $ will be:
 $ \begin{gathered}
   \Rightarrow \Delta {T_f} = 3 \times 1.86 \times 0.1 \\
   \Rightarrow 0.558 \\
\end{gathered} $
So, freezing point can be calculated by formula:
 $ {T_f} = 273.15 - \Delta {T_f} $
So, freezing point of $ N{a_2}S{O_4} $ will be:
 $ \begin{gathered}
  {T_{f(2)}} = (273.15 - 0.558)K \\
   \Rightarrow 272.592K \\
\end{gathered} $
Now, in the case of $ KCl $ ,
Degree of dissociation ( $ i $ ) = $ 2 $
So, depression in freezing point in $ N{a_2}S{O_4} $ will be:
 $ \begin{gathered}
   \Rightarrow \Delta {T_f} = 2 \times 1.86 \times 0.1 \\
   \Rightarrow 0.372 \\
\end{gathered} $
So, freezing point can be calculated by formula:
 $ {T_f} = 273.15 - \Delta {T_f} $
So, freezing point of $ N{a_2}S{O_4} $ will be:
 $ \begin{gathered}
  {T_{f(3)}} = (273.15 - 0.372)K \\
   \Rightarrow 272.778K \\
\end{gathered} $
Here we can see $ {T_{f(3)}} > {T_{f(2)}} > {T_{f(3)}} $
So, the order will be $ 0.10mol/kgKCl < 0.10mol/kgN{a_2}S{O_4} < 0.10mol/kgB{a_3}{(P{O_4})_2} $
Hence, Option B is correct.

Note
Van’t hoff factor is used to express the extent of association or dissociation of solutes in solution. It is defined as the ratio of normal molar mass to the observed molar mass of the solute. It is also the ratio of number of particles after dissociation or association to the number of particles without dissociation or association.