Question

Ethylene glycol is used as a coolant in car radiators, in order to prevent the solution from freezing at -0.3⁰C. The amount of ethylene glycol to be added to 5 Kg of water is (for water $K_f$ = 1.86 $K{m}^{-1}$):
A. 20 g
B. 50 g
C. 40 g
D. 30 g

Answer 1

int: To solve this question you have to use the formula of Depression in freezing point and it is given as $\mathrm{\Delta }{T}_f=K_f\times m$ or it can be written as $\mathrm{\Delta }{T}_f=K_f\times {\displaystyle \frac{molesofsolute}{massofsolventinkg}}$ where, $\mathrm{\Delta }{T}_f$ is the change in freezing point, $K_f$is the freezing point constant and m is the molality. Change the unit of weight from Kg to gram.

Complete step by step solution:
Ethylene glycol is a chemical which is used in many industrial and commercial applications and it also acts as an antifreeze and coolant in automobiles. When ethylene glycol is added in the water it reduces its freezing point and protects the solution from freezing in cold areas.
From your chemistry lessons you have learnt about the freezing point and depression is freezing point. Freezing point of any matter is defined as a temperature at which a liquid tends to turn into a solid when it is cooled. Depression in freezing point is the colligative property which is seen in the solution as an outcome when solute molecules are introduced to a solvent . The freezing point of the solutions are always less than the pure solvent and is directly proportional to the solute molality.
Depression in freezing point is given as, $\mathrm{\Delta }{T}_f=K_f\times m$ or it can be written as,
$$\mathrm{\Delta }{T}_f=K_f\times {\displaystyle \frac{molesofsolute}{massofsolvent(Kg)}}$$
$$\therefore \mathrm{\Delta }{T}_f={\displaystyle \frac{1000\times K_f\times w_1}{w_2\times M}}(moles={\displaystyle \frac{Givenmass}{molarmass}})$$ - (1)
Where, $\mathrm{\Delta }{T}_f$ = change in freezing point,
$K_f$ = freezing constant which is given as 1.86 $K{m}^{-1}$
$w_1$ = mass of ethylene glycol
$w_2$= mass of water (Solvent) which is 5kg = 5000g
and M = molar mass of ethylene glycol (62 g/mol)
As we know that $\mathrm{\Delta }{T}_f=T_f^{{}^{\circ }}-T_f$
Where, $T_f^{{}^{\circ }}$ is the freezing point of water which is 0⁰C and $T_f$ is the freezing point of ethylene glycol solution which is given as -0.3⁰C
Therefore, $\mathrm{\Delta }{T}_f=0^{\circ }C-(-{0.3}^{{}^{\circ }}C)={0.3}^{{}^{\circ }}C$
Here we have to find the mass of ethylene glycol that is $w_1$
Now put all the values in equation (1), we get
$$0.3={\displaystyle \frac{1.86\times 1000\times w_1}{62\times 4000}}$$
Hence, $w_1=50g$

Thus the correct option will be (B).

Note: Colligative properties are the properties of solutions which rely on the proportion that is the number of solute particles to the number of solvent molecules in a given solution. In the question you have to change the unit of weight which is given in Kg to g. Change in freezing point is equal to the freezing point of solvent – freezing of solute.