Question

Addition of 0.643g of a compound to 50mL benzene (Density of benzene $ = 0.879{{gc}}{{{m}}^{ - 3}}$ ) lowers the freezing point from ${5.51^ \circ }{{C}}$ to ${5.01^ \circ }{{C}}$. If ${{{K}}_{{f}}}$ is $5.12$, calculate the molar mass of the compound.

Answer 1

Hint: The freezing point is the temperature at which a substance in liquid state changes to solid-state. Theoretically, when the vapor pressure of a liquid equals the vapor pressure of a solid. Moreover, the freezing point depends on the van’t Hoff factor and the molality.

Complete step by step answer:
The vapor pressure of a solution is decreased when a non-volatile solute is added to a volatile solvent. This decreases vapor pressure has many properties. Freezing point depression is denoted by $\Delta {{{T}}_{{f}}}$.
\[\Delta {{{T}}_{{f}}} = {{{K}}_{{f}}}{{m}}\], where \[{{{K}}_{{f}}}\] is the freezing point depression constant.
\[{{m}}\]is the molality of the solution.
It is given that the mass of solute, ${{{m}}_{{s}}} = 0.643{{g}}$
The volume of benzene, ${{{V}}_{{b}}} = 50{{mL}}$
Density of benzene, ${\rho _{{b}}} = 0.879{{gc}}{{{m}}^{ - 3}}$
Freezing point depression constant, ${{{K}}_{{f}}} = 5.12$
Freezing point depression, $\Delta {{{T}}_{{f}}} = {5.51^ \circ }{{C}} - {5.01^ \circ }{{C = 0}}{{.5}}{{{0}}^ \circ }{{C}}$
Mass of benzene can be calculated by multiplying density with volume of benzene.
Mass of benzene, ${{{m}}_{{b}}} = {\rho _{{b}}} \times {{{V}}_{{b}}} = 0.879{{gc}}{{{m}}^{ - 3}} \times 50{{c}}{{{m}}^3} = 43.95{{g}}$
Molality can be calculated by dividing the number of moles of solute in ${{kg}}$ of solvent. The number of moles is the mass of solute divided by the molar mass of solute.
Combining the above definitions, we get the formula of molality.
Molality, ${{m = }}\dfrac{{{{{m}}_{{s}}}}}{{{{{M}}_{{s}}} \times {{{m}}_{{b}}}}} \times 1000$, where ${{{M}}_{{s}}}$ is the molar mass of solute.
Substituting the values, we get
 ${{m = }}\dfrac{{0.643g}}{{{{{M}}_{{s}}} \times 43.95{{g}}}} \times 1000$
Substituting the value in the formula of freezing point depression, we get
\[{0.50^ \circ }{{C}} = 5.12 \times \dfrac{{0.643{{g}} \times 1000}}{{{{{M}}_{{s}}} \times 43.95{{g}}}}\]
On simplification, we get
\[{0.50^ \circ }{{C}} = \dfrac{{3292.6}}{{{{{M}}_{{s}}} \times 43.95{{g}}}} = \dfrac{{74.91}}{{{{{M}}_{{s}}}}}\]

Thus, molar mass of solute, ${{{M}}_{{s}}} = \dfrac{{74.91}}{{0.50}} = 149.8{{g}}.{{mo}}{{{l}}^{ - 1}}$.

Note: When solutes are added to a solvent, forming a solution, solute molecules disrupt the formation of solvent’s crystals. This disruption in the freezing process results in a depression of freezing point for the solution compared to the solvent.