Question

What will be the molecular weight of \[{\text{CaC}}{{\text{l}}_2}\] determined in its aq. solution experimentally from depression of freezing point?
(A) 111
(B) < 111
(C) > 111
(D) data insufficient

Answer 1

Hint: This question can be solved by using the concept of van't hoff factor.
The van't Hoff factor is the ratio of the observed number of moles to the theoretical number of moles of solute.
\[{\text{i = }}\dfrac{{{\text{n}}\left( {{\text{observed}}} \right)}}{{{\text{n}}\left( {{\text{theoretical}}} \right)}}\]
Van't Hoff factor is also given as the ratio of theoretical molecular weight to observed molecular weight.
\[{\text{i = }}\dfrac{{{\text{M}}\left( {{\text{theoretical}}} \right)}}{{{\text{M}}\left( {{\text{observed}}} \right)}}\]

Step by step answer: In aqueous solution, \[{\text{CaC}}{{\text{l}}_2}\] undergoes ionization to form calcium cations and chloride anions.

\[{\text{CaC}}{{\text{l}}_2} \to {\text{C}}{{\text{a}}^{2 + }}{\text{ + 2C}}{{\text{l}}^ - }\]

Thus, the ionization of one molecule of \[{\text{CaC}}{{\text{l}}_2}\] gives three ions. Hence, the vant Hoff factor for \[{\text{CaC}}{{\text{l}}_2}\] is 3.
\[{\text{i = }}\dfrac{{{\text{n}}\left( {{\text{observed}}} \right)}}{{{\text{n}}\left( {{\text{theoretical}}} \right)}} = \dfrac{{1 + 2}}{1} = 3\]
The atomic weight of calcium and chlorine are 40 and 35.5 respectively. Calculate the molecular weight of \[{\text{CaC}}{{\text{l}}_2}\] .
\[40 + 2\left( {35.5} \right) = 40 + 71 = 111{\text{ g/mol}}\]
For \[{\text{CaC}}{{\text{l}}_2}\] the theoretical molecular weight is 111. This is same as calculated above.
Substitute the values in the expression of the vant Hoff factor in terms of molecular weight and calculate the observed molecular weight:
\[
  {\text{i = }}\dfrac{{{\text{M}}\left( {{\text{theoretical}}} \right)}}{{{\text{M}}\left( {{\text{observed}}} \right)}} \\
 \Rightarrow {\text{3 = }}\dfrac{{{\text{111}}}}{{{\text{M}}\left( {{\text{observed}}} \right)}} \\
 \Rightarrow {\text{M}}\left( {{\text{observed}}} \right) = \dfrac{{111}}{3} < 111 \\
 \]
Thus, the molecular weight of \[{\text{CaC}}{{\text{l}}_2}\] determined in its aqueous solution experimentally from depression of freezing point will be less than 111.

Hence, the option (B) is the correct option.

Note: The depression in the freezing point is the colligative property. It depends on the number of solute particles and is independent of the nature of solute particles. In some solvents, certain solutes undergo association or dissociation. For such situations, the van't Hoff factor is used.