Question

Which of the following has the highest freezing point?
A.0.1 m sucrose
B.0.01 m NaCl
C.0.1 m NaCl
D.0.01m $$N{a}_{2}S{O}_4$$

Answer 1

Hint: To solve this question, we must understand the factors by which the freezing point of a solution is affected. The freezing point of a solution is directly proportional to the concentration of the solute as well the nature of the solute (whether it is ionic or covalent in nature)

Formula Used: $$\mathrm{\Delta }{T}_f=i(K_f.m)$$

Complete Step-by-Step Answer:

Before we move forward with the solution of the given question, let us first understand some important basic concepts.
While calculating the values for physical properties of solutions like freezing point, boiling point, etc, many a times, the compound does not dissociate into smaller ions. Hence, we consider the general case where we consider the solution to be one single unit. But there are ionic solutions which involve the dissociation into ions. Here, each ion must be accounted for because it too contributes to the variation of these physical properties. To rectify this situation, we use a correction factor known as the Van’t Hoff factor.
The freezing point of the given solutions can be compared on the basis of their concentration values and their Van’t Hoff factors. The mathematical formula for calculating the freezing point can be given as:
 $$\mathrm{\Delta }{T}_f=i(K_f.m)$$
Where, $$\mathrm{\Delta }{T}_f$$ is the change in the freezing point, $$K_f$$ is the cryogenic constant, i is the Van’t Hoff factor and m I the molality of the solution. Using this formula, we get:
1)0.1 m sucrose
Sucrose is a non – ionic compound, hence i = 1.
 $$\mathrm{\Delta }{T}_f$$ = (1) (0.1 $$K_f$$ ) = 0.1 $$K_f$$
2)0.01 m NaCl
NaCl is an ionic compound with 2 ions. Hence, i = 2
 $$\mathrm{\Delta }{T}_f$$ = (2) (0.01 $$K_f$$ ) = 0.02 $$K_f$$
3)0.1 m NaCl
NaCl is an ionic compound with 2 ions. Hence, i = 2
 $$\mathrm{\Delta }{T}_f$$ = (2) (0.1 $$K_f$$ ) = 0.2 $$K_f$$
4)0.01m $$N{a}_{2}S{O}_4$$
$$N{a}_{2}S{O}_4$$ is an ionic compound with 3 ions. Hence, i = 3
 $$\mathrm{\Delta }{T}_f$$ = (3) (0.01 $$K_f$$ ) = 0.03 $$K_f$$
Hence, the highest freezing point is 0.1 m of NaCl solution.

Hence, Option C is the correct option

Note: Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative properties.