Question

For a solution of 0.849 g of mercurous chloride in 50 g of $HgC{{l}_{2}}$(l), the freezing point depression is ${{1.24}^{\circ }}C$. ${{K}_{f}}$ for $HgC{{l}_{2}}$ is 34.3. What is the state of mercurous chloride in $HgC{{l}_{2}}$?
[ Hg – 200 g, Cl – 35.5 g]
(a)- As $H{{g}_{2}}C{{l}_{2}}$ molecules
(b)- As HgCl molecule
(c)- As $H{{g}^{+}}$ and $C{{l}^{-}}$ ions
(d)- As $Hg_{2}^{2+}$ and $C{{l}^{-}}$ ions

Answer 1

Hint: The formula of mercurous chloride is $H{{g}_{2}}C{{l}_{2}}$. The formula that can be used to solve the above question is the expression of depression in freezing point, it states that $\Delta {{T}_{f}}={{K}_{f}}\text{ x m x i}$, where $\Delta {{T}_{f}}$ is depression in freezing point, ${{K}_{f}}$ is cryoscopic constant, m is the molality of the solution, and i is the association or dissociation power.

Complete answer:
The given solution in the question is of mercurous chloride dissolved in $HgC{{l}_{2}}$, so the solute is mercurous chloride and the solvent is $HgC{{l}_{2}}$. The formula of mercurous chloride is $H{{g}_{2}}C{{l}_{2}}$. So, we can calculate the Molecular mass of mercurous chloride by the information given in the question as:
$H{{g}_{2}}C{{l}_{2}}=2\text{ x 200 + 2 x 35}\text{.5 = 471}$
The formula that can be used to solve the above question is the expression of depression in freezing point, it states that $\Delta {{T}_{f}}={{K}_{f}}\text{ x m x i}$
Where $\Delta {{T}_{f}}$ is depression in freezing point, ${{K}_{f}}$ is cryoscopic constant, m is the molality of the solution, and i is the association or dissociation power. By finding the value of i we can find the state of mercurous chloride.
$\Delta {{T}_{f}}$ and ${{K}_{f}}$ are ${{1.24}^{\circ }}C$and 34.3 respectively. Molality will be:
$Molality=\dfrac{\dfrac{0.849}{417}}{\dfrac{50}{1000}}$
$Molality=\dfrac{0.849\text{ x 1000}}{417\text{ x 50}}=0.040\text{ m}$
Now, putting the values we can calculate the i as:
$1.24=34.3\text{ x 0}\text{.04 x i}$
On solving this, we get the value of i around 1 which means there is no dissociation or association of mercurous chloride ions in the solution. Hence is found as $H{{g}_{2}}C{{l}_{2}}$ molecules.

Therefore, the correct answer is an option (a).

Note:
If the value of i was less than one then the molecules will combine together or there is an association of molecule, and if the value of i is greater than 1 then there is splitting into ions or dissociation of molecules.