Question

A solution containing of $\text{ ZnC}{{\text{l}}_{\text{2}}}\text{ }$ in $\text{ }125.0\text{ }g\text{ }$ of water freezes at$\text{ }-0.23{}^\circ \text{C }$ . The apparent degree of dissociation of the salt is: $\text{ }{{\text{K}}_{\text{f}}}\text{ }$ for water $\text{ 1}\text{.86 K kg mo}{{\text{l}}^{-1}}\text{ }$ , atomic mass; $\text{ Zn = 65}\text{.3 }$ and$\text{ Cl = 35}\text{.5 }$ )
A) $\text{ 1}\text{.36}{\scriptstyle{}^{0}/{}_{0}}\text{ }$
B) $\text{ 2}\text{.47}{\scriptstyle{}^{0}/{}_{0}}\text{ }$
C) $\text{ 73}\text{.5}{\scriptstyle{}^{0}/{}_{0}}\text{ }$
D) $\text{ 7}\text{.35}{\scriptstyle{}^{0}/{}_{0}}\text{ }$

Answer 1

Hint: The elevation of the depression in freezing point $\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{f}}}\text{ }$ is a colligative property. It depends on the amount of solute. If solute dissociates into the solution, then dissociation constant for general salute $\text{ AB }$ is given as,
$\text{ }\begin{matrix}
   {} & \text{AB} & \rightleftharpoons & {{\text{A}}^{\text{+}}} & + & {{\text{B}}^{-}} \\
   \text{Before dissociation} & \alpha & {} & 0 & {} & 0 \\
   \text{After dissociation} & (1-\alpha )C & {} & C\alpha & {} & C\alpha \\
\end{matrix}$
The degree of dissociation is related to the number of moles after dissociation. That is,
$\text{ i = }\dfrac{\text{Normal molar mass}}{\text{ Observed molar mass}}\text{ = }\dfrac{1+\alpha }{1}\text{ }$

Complete answer:
The freezing point, $\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{f}}}\text{ }$of a liquid, is the temperature at which the liquid exists in equilibrium between the solid and the liquid so that the vapour pressure of the solid phase and liquid phase are equal.
When a non-volatile solute is added to a liquid, the escaping tendency forms liquid to solid disease but from solid to liquid is unaffected, thus solid starts to dissolve. To prevent this, the lowering in temperature is done known as the depression in the freezing point.
 The inorganic salts may associate or dissociate into the solution. Therefore, the elevation is the boiling point is related to the van’t Hoff factor, molality of the solution, and ebullioscopic constant as follows,
$\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{f}}}\text{ = i m }{{\text{K}}_{\text{f}}}\text{ }$
We are given the following data:
The cryoscopic constant of the solution is\[\text{ }{{\text{K}}_{\text{f}}}\text{ = 1}\text{.86 kg Kmo}{{\text{l}}^{\text{-1}}}\text{ }\],
Weight of the compound $\text{ ZnC}{{\text{l}}_{\text{2}}}\text{ }$ given,${{w}_{1}}\text{ = 0}\text{.85 gm }$
Weight of water (solvent), $\text{ }{{w}_{2\text{ }}}=125.0\text{ g }$
Freezing point difference $\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{f}}}\text{ }$is,$\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{f}}}\text{ }=\text{ 0}\text{.2}{{\text{3}}^{\text{0}}}\text{C }$
The molecular weight $\text{ ZnC}{{\text{l}}_{\text{2}}}\text{ }$is $\text{ MW of ZnC}{{\text{l}}_{\text{2}}}\text{ = 1}\times 65.5\text{ + 2}\times \text{ 35}\text{.5 = 136}\text{.5 }$
We are interested to find the degree of dissociation of $\text{ ZnC}{{\text{l}}_{\text{2}}}\text{ }$
We will solve this question by considering the van’t Hoff factor ‘i’.
Let's first calculate the molality of $\text{ ZnC}{{\text{l}}_{\text{2}}}\text{ }$.
$\text{ molality = }\dfrac{{{w}_{2}}\times 1000}{{{w}_{1}}\times {{\text{M}}_{\text{2}}}}=\dfrac{0.85\times 1000}{125\times 136.3}=0.05$
Then the value of the van’t Hoff factor is as follows,
\[\text{ i = }\dfrac{\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{f}}}}{\text{m }{{\text{K}}_{\text{f}}}\text{ }}\text{ = }\dfrac{0.23}{\left( 0.0498 \right)\times \left( 1.86 \right)}\text{ = 2}\text{.478 }\]
Thus, the van’t Hoff factor is equal to $\text{ 2}\text{.47 }$ .
The $\text{ ZnC}{{\text{l}}_{\text{2}}}\text{ }$dissociated into its ions in the solvent. The dissociation of $\text{ ZnC}{{\text{l}}_{\text{2}}}\text{ }$forms a $\text{ Z}{{\text{n}}^{\text{2+}}}\text{ }$and two $\text{ C}{{\text{l}}^{-}}\text{ }$ ions. The reaction is as follows,$\text{ }\begin{matrix}
   {} & \text{ZnC}{{\text{l}}_{\text{2}}} & \rightleftharpoons & \text{Z}{{\text{n}}^{\text{2+}}} & + & \text{2C}{{\text{l}}^{-}} \\
   \text{Before dissociation} & \alpha & {} & 0 & {} & 0 \\
   \text{After dissociation} & (1-\alpha )C & {} & C\alpha & {} & 2C\alpha \\
\end{matrix}$
 Where, $\text{ }\alpha \text{ }$ is the degree of dissociation and C is the concentration of the solution.
Thus the total number of moles after the dissociation is equal to,
 $\text{ total no}\text{.of moles = 1 }-\alpha \text{ +}\alpha \text{ + 2}\alpha \text{ = 1+ 2}\alpha \text{ }$
 Hence, the degree of dissociation would be calculated from,
$\text{ i = }\dfrac{\text{Normal molar mass}}{\text{Observed molar mass}}\text{ = }\dfrac{1\text{ + 2}\alpha }{1}\text{ }$
Substitute the values of van’t Hoff factor to get the degree of dissociation,
 $\begin{align}
  & \text{ i = 2}\text{.47 = }1\text{ + 2}\alpha \\
 & \Rightarrow \text{ }\alpha \text{ = }\dfrac{\text{2}\text{.47}-1}{2} \\
 & \therefore \alpha \text{ = 0}\text{.739 } \\
\end{align}$
Thus the degree of dissociation in the percentage is equal to,
 $\text{ }\alpha {\scriptstyle{}^{0}/{}_{0}}\text{ = 0}\text{.735 }\times \text{ 100}{\scriptstyle{}^{0}/{}_{0}}\text{ = 73}\text{.5}{\scriptstyle{}^{0}/{}_{0}}\text{ }$
Therefore, degree of dissociation of $\text{ ZnC}{{\text{l}}_{\text{2}}}\text{ }$ is equal to $\text{ 73}\text{.5}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ .

Hence, (C) is the correct option.

Note:
The inorganic salt like $\text{ ZnC}{{\text{l}}_{\text{2}}}\text{ }$ dissociates into the solution. In such a case, the number of effective particles increases, and therefore, the colligative properties like osmotic pressure, depression in freezing point, and elevation in boiling point are much higher than those of the calculated for the undissociated molecule. Note that, the van't Hoff factor for dissociation constant is always greater than 1 and for association, the van’t Hoff’s factor is less than one.