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Question: The value of \({ K }_{ f }\) of water is \({ 1.86K\quad kg mol }^{ -1 }\). If \({ 1 }\) kg of water ...

The value of Kf{ K }_{ f } of water is 1.86Kkgmol−1{ 1.86K\quad kg mol }^{ -1 }. If 1{ 1 } kg of water is filled in your automobile radiation then how much gm of ethylene glycol will be added to decrease the freezing point of the solution to −2.8∘C{ -2.8 }^{ \circ }{ C }.

Explanation

Solution

Hint: Freezing point:- The temperature at which liquid turns into a solid when cooled is called a freezing point of that particular matter.

Complete answer:
Depression in freezing point:- When we add a non-volatile solid, in the pure liquid, the freezing point of the liquid is lowered, as the vapour pressure of the liquid is lowered, due to the presence of non-volatile solute particles at the surface of the liquid.
It is given that,
Let WA{ W }_{ A } be the mass of ethylene glycol needed to be added to the automobile radiation.
Mass of the solvent (water) = WA=1kg=1000g{ W }_{ A }= { 1 }kg = { 1000g }
Kf{ K }_{ f } of water = 1.86Kkgmol−1{ 1.86K\quad kg mol }^{ -1 }.
Freezing point of water, Tf=0∘C{ T }_{ f }= { 0 }^{ \circ }{ C }
Freezing point of solution = −2.8∘C{ -2.8 }^{ \circ }{ C }
Mass of the solute, MB=62g/mol{ M }_{ B }= { 62 }g/mol
Now, use the formula;
ΔTf=Tf−Ts=1000×Kf×WB÷WA×MB\Delta { T }_{ f }{ =T }_{ f }{ -T }_{ s }{ =1000\times K }_{ f }{ \times W }_{ B }\div { W }_{ A }{ \times M }_{ B }

WB=(Tf−Ts)×WA×MB÷1000×Kf{ W }_{ B }{ =(T }_{ f }{ -T }_{ s }{ )\times W }_{ A }{ \times M }_{ B }{ \div 1000\times K }_{ f }
= 0−(−2.8)×1000×62÷1000×1.86{ \\{ 0-(-2.8)\\} \times 1000\times 62\div 1000\times 1.86 }
= 280÷3{ 280\div 3 }
= 93.33g{ 93.33g }
Therefore, the mass of ethylene glycol added, WB=93.33g{ W }_{ B }= { 93.33g }.

Additional Information:
Adding a solute to a solvent decreases the temperature at which the liquid solvent becomes a solid.
At the freezing point, the vapour pressure of both the solid and liquid form of a compound must be equal.
It is a colligative property.

Note: The possibility to make a mistake is that there is a difference between the freezing point and freezing point depression. Freezing point is the temperature at which liquid turns into a solid when cooled is called a freezing point of that particular matter while freezing point depression is the difference in the freezing points of the solution.