Question

$1\% $ aqueous solution of \[{\text{Ca}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\] has freezing point:
A. \[{0^ \circ }{\text{C}}\]
B. less than \[{0^ \circ }{\text{C}}\]
C. \[{1^ \circ }{\text{C}}\]
D. \[{2^ \circ }{\text{C}}\]

Answer 1

Hint:Freezing point depression is related with molality. Molality changes with temperature. Amount of change in freezing point depends upon the number of particles dissolved and not the type.
Given data:
Mass by volume percentage of \[{\text{Ca}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\] aqueous solution \[ = 1\% \]

Complete step by step answer:Solutions have different properties than either the solute or solvent used to make the solution. These properties are divided into colligative and non-colligative properties.
Colligative properties depend only on the number of dissolved particles in their solution, not their identity. They are of four types.
Freezing point depression is one of its types. It discusses the difference in temperature between the freezing point of a solution and that of the pure solvent. Freezing point is the temperature at which solid and liquid are in equilibrium under \[1{\text{atm}}\]. Addition of solute decreases the vapor pressure and thereby decreases the freezing point.
We can calculate the freezing point depression using the following formula:
\[\Delta {{\text{T}}_{\text{f}}} = - {\text{i}}{{\text{K}}_{\text{f}}}{\text{m}}\], where \[\Delta {{\text{T}}_{\text{f}}}\]is the freezing point depression.
\[{\text{i}} \to \]van’t Hoff factor
\[{{\text{K}}_{\text{f}}} \to \]molal depression constant
\[{\text{m}} \to \]molality
Molecular weight of \[{\text{Ca}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\]is equal to the atomic masses of one calcium, two nitrogen and six oxygen.
i.e. molecular weight of \[{\text{Ca}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\], \[{\text{W}} = 40.078 + \left( {2 \times 14.007} \right) + \left( {6 \times 15.999} \right) = 164.086 \sim 164.10{\text{gmo}}{{\text{l}}^{ - 1}}\]
Molal concentration, \[{\text{m}}\], is the number of moles of solute in kg of solvent.
Mass by volume percentage of \[{\text{Ca}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\] aqueous solution\[ = 1\% \]
i.e. the solution contains \[1{\text{g}}\]\[{\text{Ca}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\]in \[100{\text{mL}}\]solution or \[10{\text{g}}\]\[{\text{Ca}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\]in \[1000{\text{mL}}\]or \[1{\text{L}}\]solution.
Therefore weight, \[{\text{w}} = 10{\text{g}}\]
Molal concentration, \[{\text{m}} = \dfrac{{\text{w}}}{{{\text{W}} \times {\text{V}}}} = \dfrac{{10{\text{g}}}}{{164.10{\text{gmo}}{{\text{l}}^{ - 1}} \times 1{\text{kg}}}} = 0.061{\text{molk}}{{\text{g}}^{ - 1}}\]
Van’t Hoff factor of \[{\text{Ca}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\] is three since it dissociates into one mole of calcium and 2 moles of nitrate.
\[\therefore {\text{i}} = 3\]
Molal depression constant \[{{\text{K}}_{\text{f}}} = 1.86{\text{kgmo}}{{\text{l}}^{ - 1}}\]
Substituting all these values in the expression of freezing point depression, we get
\[
  \Delta {{\text{T}}_f} = - 3 \times 1.86{\text{kgmo}}{{\text{l}}^{ - 1}} \times 0.061{\text{molk}}{{\text{g}}^{ - 1}} \\
  \Delta {{\text{T}}_f} = - {0.34^ \circ }{\text{C}} \\
\]
Hence the temperature is less than \[{0^ \circ }{\text{C}}\].

Therefore the option B is correct.

Additional information:
Negative sign is used in the equation because the freezing point of solution is less than that of pure solvent. Freezing point depression and boiling point elevation are direct results of lowering vapor pressure.

Note: In order for a liquid to freeze, it must achieve a very ordered state that results in the formation of a crystal. A solution is more difficult to freeze than a pure solvent. So a lower temperature is required to freeze the liquid.