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Question: What is the Van’t Hoff factor of \(N{{a}_{3}}P{{O}_{4}}\) in a 0.40 m solution whose freezing point ...

What is the Van’t Hoff factor of Na3PO4N{{a}_{3}}P{{O}_{4}} in a 0.40 m solution whose freezing point is -2.6C^{\circ }C.

Explanation

Solution

Before answering this question, we should first know about the Van’t Hoff factor. The ratio between the number of particles of solute and the number of particles that are produced in a solution after solute is being dissolved in the solution is the Van't Hoff factor.

Complete answer:
Van’t Hoff factor of Na3PO4N{{a}_{3}}P{{O}_{4}}is-
Generally, we will assume the value of Van’t Hoff factor of Na3PO4N{{a}_{3}}P{{O}_{4}} to be 4
The number of ions that are produced per formula unit of solute in Ionic compounds.
When Sodium phosphate is kept in an aqueous solution, It dissociates and produces positive ions i.e Sodium cations (Na+N{{a}^{+}}) and negative ions i.e Phosphate anions (PO43PO_{4}^{3-}).
Na3PO4(aq)3Na+(aq)+PO43(aq)N{{a}_{3}}P{{O}_{4}}(aq)\to 3N{{a}^{+}}(aq)+PO_{4}^{3-}(aq)
As 4 ions are produced by every formula unit of sodium phosphate so we assume the van’t Hoff factor to be 4 but that’s not the case.
Freezing point depression equation is :
ΔTf=i.Kf.b\Delta {{T}_{f}}=i.\,{{K}_{f}}.\,b
In which,
ΔTf\Delta {{T}_{f}}- freezing point depression
i – Van’t Hoff factor
Kf{{K}_{f}}- cryoscopic constant of the solvent
b – molality of the solution.
The cryoscopic constant of water is 1.86 CKgmol1^{\circ }C\,Kg\,mo{{l}^{-}}^{1}.
The freezing point depression can be seen as –
ΔTf=TfTf\Delta {{T}_{f}}=T_{f}^{\circ }-{{T}_{f}}
In which,
TfT_{f}^{\circ }- the freezing point of the pure solvent
Tf{{T}_{f}}- the freezing point of the solution
In this : ΔTf=0C(2.6C)=2.6C\Delta {{T}_{f}}={{0}^{\circ }}C-(-{{2.6}^{\circ }}C)={{2.6}^{\circ }}C
Now, we put all the values in the formula to obtain the result
ΔTf=i.Kf.b\Delta {{T}_{f}}=i.\,{{K}_{f}}.\,b
i=ΔTfKf.bi=\dfrac{\Delta {{T}_{f}}}{{{K}_{f}}.\,b}
i=2.6C1.86CKgmol1×0.40molKg1i=\dfrac{{{2.6}^{\circ }}C}{{{1.86}^{\circ }}\,C\,Kg\,mo{{l}^{-1}}\times \,0.40\,mol\,K{{g}^{-1}}}
i = 3.5
Our expected value of Van’t Hoff factor was 4 but the observed value of Van’t Hoff factor that is obtained is 3.5

Note:
The expected Van’t Hoff factor will always be greater than the observed Van’t Hoff factor. It occurs because some ions form solvation cells by binding. This is the reason the expected value is greater than the number of ions produced per formula unit.