Question

A $0.001$ molal solution of a complex $[M{A_8}]$ in water has the freezing point $- {0.0054^0}C$. Assuming $100\%$ ionization of complex salt and ${K_f}$ for ${H_2}O = 1.86K.{m^{ - 1}}$, write the correct representation for the complex.
A.$[M{A_8}]$
B.$[M{A_7}]A$
C.$[M{A_6}]{A_2}$
D.$[M{A_5}]{A_3}$

Answer 1

In the above question we have given the complex $[M{A_8}]$. For the given complex we will assume that there is $100\%$ the ionization of complex salt which means charged atoms will convert or dissociate in charged ions. So we will try to find the number of dissociated ions.
Formula Used:
Depression in Freezing point
$\Delta {T_f} = i \times {K_f} \times m$
Where $\Delta {T_f}$ represents the depression in freezing point,
${K_f}$ represents molal depression constant,
$i$ represents Vant’t Hoff factor,
$m$ represents the molality of the solution.

Complete step by step answer:
First, we will write the given quantities from the question. In the question, we have given that $m = 0.001,\Delta {T_f} = 0.0054,{K_f} = 1.86$
Here we can note that the value of freezing point depression is taken as positive $\Delta {T_f} = 0.0054$ because the negative sign mentioned in the question represents the depression or decrease in temperature.
Now we will calculate the value of the Van’t Hoff factor $i$. To calculate the value of the Van’t Hoff factor $i$ we will use the formula,
$\Delta {T_f} = i \times {K_f} \times m$ $- \left( 1 \right)$
Now substituting the given values $m = 0.001,\Delta {T_f} = 0.0054,{K_f} = 1.86$ in the above equation. After substitution, we will get,
$\Rightarrow 0.0054 = i \times 1.86 \times 0.001$
$\Rightarrow i = \dfrac{{0.0054}}{{1.86 \times 0.001}}$
$\Rightarrow i = 2.9 \approx 3$
In the above step, we get the value of $i$. This value represents the total number of dissolved particles the solute produced. So here we got the value $i = 3$ which means the number of particles the solute produced is $3$.
Simply one molecule of the complex on dissociation gives $3$ particles. Now we will search the complex from the options which will give $3$ particles on dissociation.
The correct representation of complex is $[M{A_6}]{A_2}$ which will give $3$ particles on dissociation.
Therefore, the correct option is (C).

Note:
-It is to be noted that the value of the Van’t Hoff factor $i$ is always positive.
-If the value $i$ is greater than $1$, there will be the dissociation of particles.
-If the value $i$ is less than $1$, there will be an association of particles.