Question

The addition of $0.643{\text{g}}$ of a compound to $50{\text{mL}}$ of benzene $\left( {{\text{density}} = 0.879{\text{g m}}{{\text{L}}^{ - 1}}} \right)$ lowers the freezing point from $5.51$ to ${5.03^o}{\text{C}}$ . If ${{\text{K}}_{\text{f}}}$ for benzene is $5.12{\text{K}}/{\text{m}}$ , calculate the molecular weight of compound.

Answer 1

Depression in freezing point is the property of decrease in freezing point of the solution when some non volatile solute is dissolved with the solvent. ${{\text{K}}_{\text{f}}}$ is the intensive property of solvent like water and does not depend upon solute or solution. We shall substitute the values given in the formula.
Formula Used:
$\Delta {{\text{T}}_{\text{f}}} = {{\text{K}}_{\text{f}}}{\text{m}}$ is used to determine depression in freezing point of solution where m is molality of that solution.

Complete step by step answer:
Depression in freezing point is the property of decrease in freezing point of the solution when some non volatile solute is dissolved with the solvent. The depression in freezing point is given by $\Delta {{\text{T}}_{\text{f}}}$ . Freezing point is defined as the temperature at which the liquid and the solid forms of the same substance are in equilibrium and hence at this state have the same vapour pressure. As vapour pressure of the solution is less than that of pure solvent when a non volatile solute like glacial acetic acid is added to it. As freezing point is the temperature at which vapour pressure of the liquid and the solid phase are equal, therefore for the solution, this will occur at lower temperature. This is known as depression in freezing point.
Depression in freezing point can be given by $\Delta {{\text{T}}_{\text{f}}} = {\text{T}}_{\text{f}}^o - {{\text{T}}_{\text{f}}}$ where ${\text{T}}_{\text{f}}^o$ is freezing point of pure solvent and ${{\text{T}}_{\text{f}}}$ is freezing point of solution. Molal depression constant is defined as the decrease in freezing point when the molality of solution is unity. Therefore, $\Delta {{\text{T}}_{\text{f}}} = {{\text{K}}_{\text{f}}}{\text{m}}$ where ${{\text{K}}_{\text{f}}}$molal depression constant and m is is molality of solution.
As given volume of solvent or benzene is $50{\text{mL}}$ and ${\text{density}} = 0.879{\text{g m}}{{\text{L}}^{ - 1}}$ , therefore mass of solvent or benzene can be given as: ${\text{mass}} = 50 \times 0.879 = 43.95{\text{g}}$ . Molality can be given as: ${\text{Molality}} = \dfrac{{{\text{moles of solute}}}}{{{\text{mass of solvent}}\left( {{\text{in gram}}} \right)}} \times 1000 = \dfrac{{{\text{mass of solute}}}}{{{\text{molar mass of solute}} \times {\text{mass of solvent}}}} \times 1000$ . Putting values of mass of solute or a compound, mass of solvent in molality we get, ${\text{Molality}} = \dfrac{{0.643}}{{{\text{M}} \times 43.95}} \times 1000$ .
As now we have molarity and ${{\text{K}}_{\text{f}}}$ for benzene is $5.12{\text{K}}/{\text{m}}$ , depression in freezing point can be given as:
$5.51 - 5.03 = \dfrac{{0.643}}{{{\text{M}} \times 43.95}} \times 1000$

On solving the molar mass of the compound comes out to be $156{\text{g mo}}{{\text{l}}^{ - 1}}$ .

Note:
Colligative properties are those properties of a solution that depend on the number of solute particles dissolved in it and do not depend on the nature of solute particles. Relative lowering of vapour pressure, osmotic pressure, elevation in boiling point and depression in freezing point are colligative properties.