Question

For $1.00g$ of a non-electrolyte solute dissolved in $50g$ of benzene, freezing point lowered by $0.40K$ . The freezing point depression constant of benzene is \[5.12{\text{ }}Kkgmo{l^{ - 1}}\] . Find the molar mass of the solute:
A.$236gm/mol$
B.$256gm/mol$
C.$266gm/mol$
D.$274gm/mol$

Answer 1

Hint: We need to know what are colligative properties and understand the concept of depression of freezing point. The properties of solutions which depend only on the number of solute and solvent particles and not on their nature are known as colligative properties. There are four colligative properties in chemistry namely Relative lowering of vapour pressure, Elevation in boiling point, depression in freezing point and osmotic pressure.

Complete step by step answer:
We have to remember that the freezing point of a substance is the temperature at which the vapour pressure of the solid phase is equal to the vapour pressure of the liquid phase. Hence it is observed that when a non-volatile, nonelectrolyte solute is added to a solvent, the freezing point of the solvent in solution decreases. This phenomenon is known as depression in freezing point given by the symbol $\Delta {T_f}$ .
By the definition of depression in freezing point,
$\Delta {T_f} \propto m$ ,where $m$is the molality of the solution.
Or $\Delta {T_f} = {K_f}m$, where ${K_f}$ is a constant of proportionality known as freezing point depression constant.
The molality $m$is calculated as,
$m = \dfrac{{{\text{Weight of solute}}\left( {{W_b}} \right)}}{{{\text{Molar mass of solute}}\left( {{M_b}} \right) \times {\text{Weight of solvent in kg}}\left( {{W_a}} \right)}}$
Therefore, $\Delta {T_f} = {K_f}\dfrac{{{W_b}}}{{{M_b} \times {W_a}(kg)}}$
Given in the question:
${W_b} = 1.00g$
${W_a} = 50g = 50/1000kg$
$\Delta {T_f} = 0.40K$
\[{K_f} = 5.12{\text{ }}Kkgmo{l^{ - 1}}\]
We are to calculate the molar mass of the solute ${M_b}$.
We know, $\Delta {T_f} = {K_f}\dfrac{{{W_b}}}{{{M_b} \times {W_a}(kg)}}$
Now we can substitute the known given values we get,
$0.40 = 5.12\dfrac{{1.00 \times 1000}}{{{M_b} \times 50}}$
On calculating,
${M_b}$=$256gm/mole$
Hence the correct option is option (B).

Note:
We must note that the freezing point depression constant is sometimes also referred to as molal depression constant or cryoscopic constant. Numerically,it is equal to the depression in the freezing point of one molal solution and has the units $Kmola{l^{ - 1}}$ or $K{m^{ - 1}}$ or $Kmo{l^{ - 1}}kg$ .