Question

When$3.24g$ of mercuric nitrate, $Hg{(N{O_3})_2}$ is dissolved in $1000g$ of water, the freezing point of the solution is found to be $- {0.0558^0}C$ .But when $10.84g$ of $HgC{l_2}$ is dissolved in $1000g$ of water, the freezing point depression is ${0.0744^0}C$. ${K_f}$ for water is $1.86mola{l^{ - 1}}$.Now select the correct statement(s). (Given: Atomic mass of Hg $= 200$)
This question has multiple correct options.
A. Mercuric nitrate dissociates to give one $H{g^{2 + }}$ and two nitrate ions.
B. $Hg{(N{O_3})_2}$ does not dissociate.
C. $HgC{l_2}$ dissociates to give one $H{g^{2 + }}$ and two chloride ions.
D. $HgC{l_2}$ does not dissociate.

Answer 1

We will make use of our knowledge of colligative properties here. The properties which depend only upon the concentration of solute particles are called colligative properties. The four colligative properties are:
1.Lowering of vapor pressure
2.Elevation in boiling point
3.Depression in freezing point
4.Osmotic Pressure
Depression in freezing point is also a colligative property which means that it is dependent on the presence of dissolved particles and their numbers but not on their identity. When solute molecules are introduced in a solvent the freezing point of the solution lowers as compared to the pure solvent and it is directly proportional to the molality of the solute.

Complete step by step answer:
Depression in freezing point is given by the formula:

$\Delta {T_f} = {K_f} \times i\times m$

Where $\Delta {T_f}$ is the Depression in freezing point

${K_f}$is the freezing point depression constant.

$i$is the Van’t Hoff factor

$m$ is the molality of the solution

Now we will find the Van’t Hoff factor for both the solutions.

For mercuric nitrate:

$\Delta {T_f} = {K_f} \times i\times m$

$\Rightarrow 0 - ( - 0.0558) = i \times 1.86 \times \dfrac{1}{{100}}$ (Number of moles of solute is $\dfrac{1}{{100}}$ per kilogram of solvent)

$\Rightarrow i = \dfrac{{0.0588 \times 100}}{{1.86}}$

$\Rightarrow i = 3$

For $HgC{l_2}$

$\Delta {T_f} = {K_f} \times i\times m$

$m =$number of moles of solute/Volume of solvent in $kg$

$\Rightarrow 0.0744 = i \times 1.86 \times \dfrac{{\dfrac{{10.84}}{{271.5}}}}{{\dfrac{{1000}}{{1000}}}}$ (Number of moles of solute is $\dfrac{{10.84}}{{271.5}}$ per kilogram of solvent)

$\Rightarrow 0.0744 = i \times 1.86 \times 0.039$

$\Rightarrow i = \dfrac{{0.0744}}{{1.86 \times 0.039}}$

$\Rightarrow i = 1$

Now since we have found the Van’t Hoff factor for both the solutions we can give the number of ions formed after dissociation of both of the solutions. For mercuric nitrate, it is three which means that the number of ions formed after dissociation is three. Hence it will give one $H{g^{2 + }}$ and two nitrate ions. For $HgC{l_2}$ the Van’t Hoff factor is one which means that it does not dissociate in solution.

So, the correct answer is Option A,D .

Note: Melting point is not a colligative property. Depression in the melting point is a colligative property.The Van't Hoff factor gives us a degree of dissociation or association of a solute in a solvent.
Molality is the number of moles of solute present per kilogram of solvent.