Question

Given,$3.24g$ of $Hg{(N{O_3})_2}$ ​(molar mass $= 324$) dissolved in $1000g$ of water constitutes a solution having a freezing point of $- {0.0558^o}C$ while $21.68g$ of $HgC{l_2}$ ​ (Molar mass $= 271$) in $2000g$ of water constitutes a solution with a freezing point of $- {0.0744^o}C$. The ${K_f}$ ​for water is $1.86\dfrac{{K.Kg}}{{mol}}$ ​. About the state of ionization of these two solids in water it can be inferred that
A.$Hg{(N{O_3})_2}$ and $HgC{l_2}$ both are completely ionized
B.$Hg{(N{O_3})_2}$ is fully ionized but $HgC{l_2}$ is fully unionized
C.$Hg{(N{O_3})_2}$ and $HgC{l_2}$ both are completely unionized
D.$Hg{(N{O_3})_2}$ is fully unionized but $HgC{l_2}$ is fully ionized

Answer 1

We have to know that the mercuric nitrate is a chemical compound having the formula,$Hg{(N{O_3})_2}$ with molecular weight $324.7g/mol$. It is a toxic compound and it may be colorless or white soluble crystalline compound. When the mercuric nitrate is thermally decomposed, there is a formation of mercury, nitrogen dioxide and oxygen gas and this reaction takes place in the presence of high temperature. The mercuric chloride is a chemical compound having the molecular formula, $HgC{l_2}$. It is a white crystalline compound which is toxic to humans.

Complete answer:
The mercuric chloride, $HgC{l_2}$ is completely unionized. Hence, option (A) is incorrect.
According to the question, the weight of $Hg{(N{O_3})_2}$ is equal to $3.24g$ and the molecular weight is equal to $324g$. Weight of $HgC{l_2}$ is $21.68g$ and molecular weight of mercuric chloride is equal to $271g$.
And given dissolved $3.24g$ mercuric nitrate is dissolved in $1000g$ with depression in freezing point, {t_f}$$$$ - {0.0558^o}C and $21.68g$ is dissolved in $2000g$ with {t_f}$$$$ - {0.0744^o}C
$\Delta {T_f} = i \times m \times {K_f}$
Where, m is equal to given weight divided by molecular weight and i is Van’t Hoff factor and it is constant for each compound. Substitute the given values in the above equation.
$0.0558 = i \times \dfrac{{3.24}}{{324}} \times \dfrac{{1000}}{{1000}} \times 1.86$
By rearranging will get the value of ‘i’ of mercuric nitrate
$i = 3(100\% dissociated)$
$0.0744 = i \times \dfrac{{21.68}}{{271}} \times \dfrac{{1000}}{{2000}} \times 1.86$
On simplification we get,
$i \simeq 1(almost undissociated)$
Here, $Hg{(N{O_3})_2}$ is fully ionized but $HgC{l_2}$ is fully unionized.Hence, option (B) is correct.
Here, mercuric nitrate is completely ionized. Hence, option (C) is incorrect.
The mercuric nitrate is completely ionized and mercuric chloride is completely unionized. Hence, option (D) is incorrect.

Hence, option (B) is correct.

Note:
We must have to know that the relation between Van’t Hoff factor and degree of dissociation is explained on the basis of values of Van’t Hoff factor. If the value of the van't Hoff factor is greater than one, it indicates dissociation and if it is smaller than indicates association. And if the Van’t Hoff factor is equal to one, it is for non-electrolyte solute. The value of the Van't Hoff factor is never negative.