Question

# The freezing point of nitrobenzene is 278.8K. A 0.25 molal solution of a substance (MM. 120) in nitrobenzene has freezing points 276.8K. Calculate the molal depression constant of nitrobenzene.

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Hint:
When solute particles are added to any pure liquid or solution, they tend to separate the solvent molecules because of the interference of the solute particles. This interference causes a lowering in the freezing point of the pure solvent. This concept is known as freezing point depression.
Formula Used: $\Delta {T_f} = {K_f} \times molality$

The freezing point of any liquid is basically the temperature at which the free flowing mobility of the molecules of the liquid state is restricted, and due to such immobility of molecules, the given substance transforms from solid state to liquid state.
This depression in the freezing point can be mathematically formulated and calculated. The depression in the freezing point or the difference in temperature is directly proportional to the molality of the solute. This relation is equated by introducing a constant known as the molal freezing point depression constant. Mathematically, this can be represented as:
$\therefore {K_f} = \dfrac{{0.25}}{{\Delta {T_f}}}$
Where, $\Delta {T_f}$is the depression in the freezing point,
${K_f}$ is molal freezing point depression constant,
m is the molality of the solute.
Rearranging this equation, we get,
${K_f} = \dfrac{{\Delta {T_f}}}{m}$
${K_f} = \dfrac{{\Delta {T_f}}}{{0.25}}$
Here, $\Delta {T_f}$= 278.8K – 276.8K = 2K
Hence,
$\therefore {K_f} = \dfrac{2}{{0.25}}$
$\therefore {K_f} = 8K.kg.mo{l^{ - 1}}$

Hence, the molal depression constant of nitrobenzene is$8K.kg.mo{l^{ - 1}}$.

Note:
In case the molality of the solute is not given, then the molality can be calculated as a ratio of the number of moles of the solute to the weight of the solvent in kg.