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Question: What will be the freezing point of the 20.0% solution of glucose \({{C}_{6}}{{H}_{12}}{{O}_{6}}\) in...

What will be the freezing point of the 20.0% solution of glucose C6H12O6{{C}_{6}}{{H}_{12}}{{O}_{6}} in water?

Explanation

Solution

Before solving this question, we should know the formula to calculate the freezing point of the solution. The formula is ΔTf=Tf−T∗f=ikfmi\Delta {{T}_{f}}={{T}_{f}}-{{T}^{*}}_{f}=i{{k}_{f}}{{m}_{i}}, Now we can put the values in the formula and can easily solve the numerical.

Complete answer:
Let’s assume 20% by mass, 20% ww\dfrac{w}{w}, and not by volume
The formula for freezing point depression is:
ΔTf=Tf−T∗f=ikfmi\Delta {{T}_{f}}={{T}_{f}}-{{T}^{*}}_{f}=i{{k}_{f}}{{m}_{i}}
Here,
The freezing point of the liquid is Tf{{T}_{f}}(It can be for solvents as well as for the solutions) and the value for pure solvent is indicated by *.
The number of particles per formula unit that has been dissociated into the solution is indicated by Van’t Hoff factor i. 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.
The freezing point depression constant of water is kf{{k}_{f}}and the value of kf{{k}_{f}}is 1.86∘C^{\circ }C.
The molality =mol solutekg solvent\dfrac{mol\,solute}{kg\,solvent} of the solute glucose in the solvent is represented by mi{{m}_{i}} .
We will take water as 100 g which means 0.100 kg, so we have 20 g of glucose which will help us to find the molality.
mi=niwA{{m}_{i}}=\dfrac{{{n}_{i}}}{{{w}_{A}}}= wiMiwA\dfrac{\dfrac{{{w}_{i}}}{Mi}}{{{w}_{A}}}= 20g glucos⁡e×1mole180.154g glucos⁡e0.100kg water\dfrac{20g\,glu\cos e\times \dfrac{1mole}{180.154g\,glu\cos e}}{0.100kg\,water}=1.1102 mol/kg
Here, the van’t Hoff factor is 1, so i=1 and glucose being non-electrolyte, it does not ionize at all as it does not dissociate well.
And, we should know Tf∗T_{f}^{*}= 0∘C^{\circ }Cin the case of pure water.
That means △Tf=Tf\vartriangle {{T}_{f}}={{T}_{f}}
Now we have to find Tf{{T}_{f}}:
△Tf=−1(1.86 ∘Ckgmol)×1.1102molkg\vartriangle {{T}_{f}}=-1(1.86{{\,}^{\circ }}C\dfrac{kg}{mol})\times 1.1102\dfrac{mol}{kg}
= - 2.06∘C^{\circ }C
So, Tf{{T}_{f}}= -2.06∘C^{\circ }C

Note:
There is a difference between depression in freezing point and freezing point
The difference in the freezing point of the solution from its pure solvent is known as depression in the freezing point.
The temperature at which the state of liquid gets converted into the solid, so their vapor pressures are equal is known as the freezing point.