Question

Why is the freezing point depression of 0.1M sodium chloride is nearly twice that of the \[0.1M\] glucose solution?

Answer 1

Hint: Adding a solute decreases the freezing point depression due to the solute-solvent interactions. It is a colligative property. The amount of change in the freezing point is related to the number of particles of solute in a solution and does not depend on the chemical composition of the solute.

Complete step-by-step answer:
When a substance starts to freeze, the molecules slow down due to the decreases in temperature and mobility of ions; and the intermolecular forces start to take over. At the freezing point of a solvent, an equilibrium exists between the liquid state and the solid state of the solvent. This tells that the vapour pressures of both the liquid and the solid phase are equal.
On adding solute which is non-volatile, the vapour pressure of the solution is observed to be lower than the vapour pressure of the pure solvent. This causes the solid and the solution to reach equilibrium at lower temperatures and that causes freezing point depression.
Depression in freezing point can be calculate by using the below formula:
\[\Delta {T_f} = \dfrac{{{K_f} \times {w_B} \times 1000}}{{{w_A} \times {M_B}}}\]
Where, \[{K_f}\]= molal depression constant
\[{w_B}\]= weight of solute
\[{w_A}\]= weight of solvent
\[{M_B}\]= molar mass of the solute
From this formula, we conclude that freezing point depression of a liquid is inversely proportional to the molar mass of the solute. Therefore, the compound with higher molar mass will have lower freezing point depression and the compound with lower molar mass will have higher freezing point depression.
We already know, molar mass of sodium chloride, \[NaCl\] = 58.44 g/ml
Molar mass of glucose solution, \[{C_6}{H_{12}}{O_6}\] =\[108.156{\text{ }}g/ml\]
We see that the molar mass of glucose is almost double of \[NaCl\]. Since molar mass has an inverse relation with freezing point depression, \[NaCl\] will have just twice the freezing point depression of glucose as the concentration of both the solutes is the same.
Hence, we can say that the freezing point depression of \[0.1M\] sodium chloride is nearly twice than that of \[0.1M\] glucose solution.

Note: Another method is- when sodium chloride will dissociate into ions, sodium and chloride ions will be formed and due to formation of these two ions, their molar concentration will double up to\[0.2M\]. But glucose is nonpolar and does not dissociate into ions and will have the same concentration \[0.1M\] . As the concentration becomes twice, freezing point depression also becomes twice for \[NaCl\].