Question

If glycerol and methanol were sold at the same price in the market, which would be cheaper for preparing an antifreeze solution for the radiator of an automobile?
(A) Glycerol
(B) Methanol
(C) Both are equal
(D) None of these

Answer 1

Hint: The depression in the freezing point of a solution upon the addition of a solute can be given by
\[\Delta {T_f} = {K_f} \cdot m\]
Here, m is the molality of the solute and ${K_f}$ is the molal depression constant

Complete step by step solution:
An antifreeze solution is a solution which contains a material which lowers the freezing point of the solvent. It can be used in the radiators of automobiles in order to prevent the damage after the freezing of the solution.
- So, here we will consider glycerol and methanol as solute and antifreeze substances.
- Now, the depression in the freezing point of a solution upon the addition of a solute can be given by
\[\Delta {T_f} = {K_f} \cdot m{\text{ }}....(1)\]
Where ${K_f}$ is the molal depression constant and m is the molality of the solute.
Now, we know that molality of a solute can be given by the number of moles of solute dissolved in one kilogram of a solvent. As we put the value of molality in equation (1), we get
\[\Delta {T_f} = {K_f} \cdot \dfrac{{{\text{Moles of solute}}}}{{{\text{Weight of solvent}}}}{\text{ }}....{\text{(2)}}\]
We can write that number of moles of solute = $\dfrac{{{\text{Weight of solute}}}}{{{\text{Molecular weight of solute}}}}$
Putting the value of the number of moles of solute in equation (2), we get
\[\Delta {T_f} = {K_f} \cdot \dfrac{{{\text{Weight of solute}}}}{{{\text{Molecular weight of solute}} \times {\text{Weight of solute}}}}{\text{ }}........{\text{(3)}}\]
Now, we are given that both methanol and glycerol have the same antifreezing activity and they are used in the same amounts. So, only the change will be the molecular weight in equation (3). Let’s find the molecular weight of both of them in order to compare their anti-freezing activity.
- Molecular weight of Methanol ($C{H_3}OH$) = Atomic weight of C + Atomic weight of O + 4(Atomic weight of H)
Molecular weight of Methanol = $12 + 16 + 4(1) =$ $32$ $gmmo{l^{ - 1}}$
Molecular weight of Glycerol (${C_3}{H_9}{O_3}$) = 3(Atomic weight of C) + 9(Atomic weight of H) + 3(Atomic weight of O)
Molecular weight of Glycerol = $3(12) + 9(1) + 3(16) =$ $36+9+48 =$ $93$ $gmmo{l^{ - 1}}$
- Now, we can say that methanol has lower molecular weight than glycerol and thus the value of $\Delta {T_f}$ will be more for methanol than glycerol.
- Thus, we can say that we will be able to decrease the freezing point of water with less mass of methanol. The mass of the glycerol required will be more than methanol. So, we can say that methanol will be cheaper than glycerol.

Therefore, the correct answer is (B).

Note: Remember that the property that the antifreeze compound should possess is that it should be non-volatile. It decreases the vapour pressure of the solution and so the freezing point of the solution decreases.