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Question: The value of \({K_f}\) is \(1.86{\text{ K kg mo}}{{\text{l}}^{ - 1}}\) . If \(1{\text{ kg}}\) water ...

The value of Kf{K_f} is 1.86 K kg mol−11.86{\text{ K kg mo}}{{\text{l}}^{ - 1}} . If 1 kg1{\text{ kg}} water is held in your automobile radiator then how many grams of ethylene glycol must be added to decrease the freezing point of solution to −2.8∘C - {2.8^ \circ }C ?

Explanation

Solution

The lowering of the freezing point of solvents due to the addition of solutes is called freezing point depression. It is a colligative property of solutions and it is proportional to the molality of the added solute. The properties which depend on the concentration of solute ions or molecules but not on the identity of solute are known as colligative properties.
Formula used: ΔTf=Kf×m\Delta {T_f} = {K_f} \times m
Where ΔTf=\Delta {T_f} = freezing point depression
Kf={K_f} = cryoscopic constant
m=m = molality

Complete answer: On adding a solute that is non-volatile, the vapour pressure of the solution is found to be lesser than the vapour pressure of the pure solvent. This will cause the solid and the solution to reach equilibrium at lower temperatures and hence freezing point depression occurs.
Given Kf=1.86 K kg mol−1{K_f} = 1.86{\text{ K kg mo}}{{\text{l}}^{ - 1}}
The change in the freezing point of solution ΔTf=0−(−2.8)=2.8 K kg mol−1\Delta {T_f} = 0 - ( - 2.8) = 2.8{\text{ K kg mo}}{{\text{l}}^{ - 1}}
Mass of the solvent =1 kg = 1{\text{ kg}}
Molecular mass of the solute =62 = 62
We know that, ΔTf=Kf×m\Delta {T_f} = {K_f} \times m
⇒m=Weight of soluteMolecular mass of soluteMass of solvent(g)×1000\Rightarrow m = \dfrac{{\dfrac{{{\text{Weight of solute}}}}{{{\text{Molecular mass of solute}}}}}}{{{\text{Mass of solvent(g)}}}} \times 1000
⇒m=w621000×1000=w62\Rightarrow m = \dfrac{{\dfrac{w}{{62}}}}{{1000}} \times 1000 = \dfrac{w}{{62}}
ΔTf=Kf×m\Delta {T_f} = {K_f} \times m
2.8=1.86×w622.8 = 1.86 \times \dfrac{w}{{62}}
⇒w=62×2.81.86=93g\Rightarrow w = \dfrac{{62 \times 2.8}}{{1.86}} = 93g
Therefore, 9393 grams of ethylene glycol must be added to decrease the freezing point of solution to −2.8∘C - {2.8^ \circ }C.

Note:
There exists an equilibrium between the liquid state and solid state of the solvent at the freezing point of solvent. This means that the vapour pressures of both the liquid and the solid phases are equal. The radiator fluids used in many automobiles are usually made up of ethylene glycol and water. This helps in preventing the freezing of the radiator in cold seasons.