The value of $K_{f}$ of water is ${1.86 K k g m o l}^{- 1}$ . If $1$ 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 ${- 2.8}^{\circ} C$ .

Answer

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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

W_{A}

be the mass of ethylene glycol needed to be added to the automobile radiation.
Mass of the solvent (water) =

W_{A} = 1 k g = 1000 g

K_{f}

of water =

{1.86 K k g m o l}^{- 1}

.
Freezing point of water,

T_{f} = 0^{\circ} C

Freezing point of solution =

{- 2.8}^{\circ} C

Mass of the solute,

M_{B} = 62 g / m o l

Now, use the formula;

Δ T_{f} {= T}_{f} {- T}_{s} {= 1000 \times K}_{f} {\times W}_{B} \div W_{A} {\times M}_{B}

W_{B} {= (T}_{f} {- T}_{s} {) \times W}_{A} {\times M}_{B} {\div 1000 \times K}_{f}

{0 - (- 2.8)} \times 1000 \times 62 \div 1000 \times 1.86

280 \div 3

93.33 g

Therefore, the mass of ethylene glycol added,

W_{B} = 93.33 g

.

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.