Question

Cryoscopic constant of a liquid is:
A. Decrease in freezing point when 1 gram of solute is dissolved per kg of the solvent.
B. Decrease in the freezing point when 1 mole of solute is dissolved per kg of the solvent.
C. The elevation for 1 molar solution
D. A factor used for calculation of elevation in boiling point

Answer 1

Hint: The depression in freezing point is given by multiplying Van’t Hoff factor with the cryoscopic constant and molality. The cryoscopic constant is given by \[{K_f}\]. The molality is defined as the number of moles of solute dissolved in one kg of solvent.

Complete step by step answer:
When the substance in liquid state changes to solid state, the process is known as freezing. The temperature at which the liquid substance starts to change into solid is known as freezing point.
At the freezing point there is an equilibrium maintained between the liquid state of the compound and the solid state of the compound which shows that the vapour pressure of both the phases liquid and solid are the same.
When a non-volatile solute is added to the solution, there is a decrease in the vapor pressure of the solution as compared to the vapour pressure of solvent. As a result solid and solution reach equilibrium at lower temperature. This refers to the depression in freezing point.
The depression in freezing point is defined as the lowering of the freezing point solvents by addition of non-volatile solute.
The formula of depression in freezing point is shown below.
\[\Delta {T_f} = i \times {K_f} \times m\]
Where,
\[\Delta {T_f}\] is the depression in freezing point.
i is the Van’t Hoff factor
\[{K_f}\]is the cryoscopic constant
m is the molality
The cryoscopic constant is defined as the freezing point depression on dissolving a non-volatile solute in 1 kg of solvent.
Thus, the cryoscopic constant of a liquid decreases in the freezing point when 1 mole of solute is dissolved per kg of the solvent.

So, the correct answer is Option B.

Note: The depression in freezing point is a colligative property which is measured in terms of k.kg. \[mo{l^{ - 1}}\]. The depression in freezing point depends on the molar mass of the solute present in the solution.