Question

Calculate the amount of $CaC{l_2}$ (molar mass = 111 $gmo{l^{ - 1}}$) which must be added to 580 g of water so that freezing point lower by 2K, assuming $CaC{l_2}$ is completely dissociated:
[${K_f}$ For water = 1.86$ KKgmo{l^{ - 1}}$].

Answer 1

Hint: This question is asked from the topic of colligative properties of solutions. Whenever we add a solute particle in a solvent, then the freezing point of the solution will decrease in a certain amount. This lowering of the freezing point is one of the colligative properties.

Complete step by step answer:
We know that the colligative properties formulas are useful for finding out the molecular mass of the solute particles.
The formula for the decrease in freezing point is, $\Delta {T_f} = i \times \dfrac{{\left( {{K_f} \times {W_B} \times 1000} \right)}}{{{W_A} \times {M_B}}}$
Where $i$ is the van’t hoff factor, ${K_f}$ is cryoscopic constant, ${W_B}$ is the mass of solute particles, ${W_A}$ is the mass of solvent and ${M_B}$ is the molar mass of solute.
In the question, it is given that:
${K_f}$ for water = 1.86 $KKgmo{l^{ - 1}}$
${M_B}$ = 111 $gmo{l^{ - 1}}$
${W_A}$ = 580 g
$\Delta {T_f}$ = 2 K
We need to find out the${W_B}$. Also we need to find out the van’t hoff factor$i$. That is:
According to the dissociation of \[CaC{l_2}\],
$CaC{l_2} \to C{a^{2 + }} + 2C{l^ - }$
For this reaction, the value of van’t hoff factor $i$ = 3.
Substituting the all these values in the equation, we get
$2 = 3 \times \dfrac{{\left( {1.86 \times {W_B} \times 1000} \right)}}{{111 \times 580}}$
$ \Rightarrow {W_B} = \dfrac{{2 \times 111 \times 580}}{{1.86 \times 3 \times 1000}}$
$ \Rightarrow {W_B} = 23.07g$
So, the amount of \[CaC{l_2}\] is required to add so that the freezing point will be depressed by 2K is 23.07 g.

Note:
While we solve this type of problems, they will not give the value of the van't hoff factor. We need to find out this value by writing the dissociation reaction of solute and the difference between the number of moles of products and reactants will be the van’t hoff factor.