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Question: Assuming \( 100\% \) ionization, the increasing order of the freezing point of the solution will be:...

Assuming 100%100\% ionization, the increasing order of the freezing point of the solution will be:
A) 0.10mol/kgBa3(PO4)2<0.10mol/kgNa2SO4<0.10mol/kgKCl0.10mol/kgB{a_3}{(P{O_4})_2} < 0.10mol/kgN{a_2}S{O_4} < 0.10mol/kgKCl
B) 0.10mol/kgKCl<0.10mol/kgNa2SO4<0.10mol/kgBa3(PO4)20.10mol/kgKCl < 0.10mol/kgN{a_2}S{O_4} < 0.10mol/kgB{a_3}{(P{O_4})_2}
C) 0.10mol/kgNa2SO4<0.10mol/kgBa3(PO4)2<0.10mol/kgKCl0.10mol/kgN{a_2}S{O_4} < 0.10mol/kgB{a_3}{(P{O_4})_2} < 0.10mol/kgKCl
D) 0.10mol/kgKCl<0.10mol/kgBa3(PO4)2<0.10mol/kgNa2SO40.10mol/kgKCl < 0.10mol/kgB{a_3}{(P{O_4})_2} < 0.10mol/kgN{a_2}S{O_4}

Explanation

Solution

In order to arrange them in increasing order of their freezing point, we must calculate the Van’t Hoff factor. As here 100%100\% ionization is taking place, we have to calculate its degree of dissociation which is further used to calculate the freezing point.
Formula used
Tf=273.15−ΔTf{T_f} = 273.15 - \Delta {T_f}
And,
ΔTf=i×Kf×M\Delta {T_f} = i \times {K_f} \times M
Where, ii = degree of dissociation
Kf{K_f} = molal freezing point depression constant
MM = concentration of solution
Tf{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%100\% ionization, so firstly we have to calculate the degree of dissociation of every solution.
In case of Ba3(PO4)2B{a_3}{(P{O_4})_2} ,
Degree of dissociation ( ii ) = 55
So, the depression in freezing point will be given as:
ΔTf=i×Kf×M\Delta {T_f} = i \times {K_f} \times M
Where, ii = degree of dissociation
Kf{K_f} = molal freezing point depression constant
And MM = concentration of solution
So, depression in freezing point in Ba3(PO4)2B{a_3}{(P{O_4})_2} will be:
⇒ΔTf=5×1.86×0.1 ⇒0.93 \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:
Tf=273.15−ΔTf{T_f} = 273.15 - \Delta {T_f}
So, freezing point of Ba3(PO4)2B{a_3}{(P{O_4})_2} will be:
Tf(1)=(273.15−0.93)K ⇒272.22K \begin{gathered} {T_{f(1)}} = (273.15 - 0.93)K \\\ \Rightarrow 272.22K \\\ \end{gathered}
Now, in the case of Na2SO4N{a_2}S{O_4} ,
Degree of dissociation ( ii ) = 33
So, depression in freezing point in Na2SO4N{a_2}S{O_4} will be:
⇒ΔTf=3×1.86×0.1 ⇒0.558 \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:
Tf=273.15−ΔTf{T_f} = 273.15 - \Delta {T_f}
So, freezing point of Na2SO4N{a_2}S{O_4} will be:
Tf(2)=(273.15−0.558)K ⇒272.592K \begin{gathered} {T_{f(2)}} = (273.15 - 0.558)K \\\ \Rightarrow 272.592K \\\ \end{gathered}
Now, in the case of KClKCl ,
Degree of dissociation ( ii ) = 22
So, depression in freezing point in Na2SO4N{a_2}S{O_4} will be:
⇒ΔTf=2×1.86×0.1 ⇒0.372 \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:
Tf=273.15−ΔTf{T_f} = 273.15 - \Delta {T_f}
So, freezing point of Na2SO4N{a_2}S{O_4} will be:
Tf(3)=(273.15−0.372)K ⇒272.778K \begin{gathered} {T_{f(3)}} = (273.15 - 0.372)K \\\ \Rightarrow 272.778K \\\ \end{gathered}
Here we can see Tf(3)>Tf(2)>Tf(3){T_{f(3)}} > {T_{f(2)}} > {T_{f(3)}}
So, the order will be 0.10mol/kgKCl<0.10mol/kgNa2SO4<0.10mol/kgBa3(PO4)20.10mol/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.