Question

The value of $K_f$ of water is ${1.86Kkgmol}^{-1}$. 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 1

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) = $W_A=1kg=1000g$
$K_f$ of water = ${1.86Kkgmol}^{-1}$.
Freezing point of water, $T_f=0^{\circ }C$
Freezing point of solution = ${-2.8}^{\circ }C$
Mass of the solute, $M_B=62g/mol$
Now, use the formula;
$\mathrm{\Delta }{T}_f{=T}_f{-T}_s{=1000\times K}_f{\times W}_B÷W_A{\times M}_B$

$W_B{=(T}_f{-T}_s{)\times W}_A{\times M}_B{÷1000\times K}_f$
= $\{0-(-2.8)\}\times 1000\times 62÷1000\times 1.86$
= $280÷3$
= $93.33g$
Therefore, the mass of ethylene glycol added, $W_B=93.33g$.

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.