Question

A solution containing $0.5g$of $KCl$ dissolved in $100g$ of water and freezes at $0.24{}^\circ C$. Calculate the degree of dissociation of the salt. ( ${{K}_{f}}$ for water $=1.86{}^\circ C$ ). Atomic weights [ $K=39,Cl=35.5$ ]

Answer 1

Hint:The freezing point of a solution can be calculated by taking in account the concentration of the solution in terms of molality. Molality is the number of moles of solute present per gram of the solvent.The degree of dissociation of the ionic solutes can be calculated by using the value of van't’ Hoff’s factor, obtained by the ratio of the values of observed and calculated freezing points.

Formula used:
\[\Delta {{T}_{f}}=\dfrac{{{K}_{f}}\times 1000\times {w}}{W\times m}\]
Here the $\Delta {{T}_{f}}$ represents depression in freezing point which is equal to ${{T}_{f}}-{{T}_{f}}'$, where both of them represents final and initial temperature of the solvent respectively. $W$ weight of solvent, $w$ is the weight of solute.And $m$ symbolizes the molecular weight.

Complete step-by-step solution:Freezing point of a solution could be defined as that temperature, at which the liquid state of a solution freezes and gets converted to its solid state. Different liquids have different freezing points. The depression in freezing point is a type of colligative property, which is the phenomena which could be observed when we add solute to a solvent, and this results in, decrease of the freezing point of the solvent under consideration.
\[\Delta {{T}_{f}}=\dfrac{{{K}_{f}}\times 1000\times {w}}{W\times m}\]
Here the $\Delta {{T}_{f}}$ symbolises depression in freezing point which is equal to ${{T}_{f}}-{{T}_{f}}'$, here both of them represents final and initial temperature of the solvent respectively.
$W$ represents the weight of solvent, $w$ is the weight of solute.
And $m$ symbolizes the molecular weight.
Now we will write all the values, which are known to us.
Weight of the solute ($w$) which is potassium chloride is $0.5g$, weight of solvent ($W$) which is water is $100g$
The molecular weight ($m$) of the solute is $39+35.5=74.5g$
Now, substituting these values in the relation, we get,
\[\Delta {{T}_{f}}=\dfrac{{{K}_{f}}\times 1000\times {w}}{W\times m}=\dfrac{1.86\times 1000\times 0.5}{100\times 74.5}\]
Here the value of ${{K}_{f}}$ is given as $1.86$, and the rest of the values are substituted as it is.
Now, after solving this equation we get,
$\Delta {{T}_{{{f}_{cal}}}}=0.124{}^\circ C$
This value of freezing point depression is the calculated value, now we will write the observed value of freezing point depression, which is
$\Delta {{T}_{{{f}_{obs}}}}=0{}^\circ C+0.24{}^\circ C=0.24{}^\circ C$
Now, from these values we will calculate the van't' Hoff’s factor.
$i$ represents the van’t Hoff’s factor, which is the ratio of observed value and the calculated value.
So, we get,
$i=\dfrac{\Delta {{T}_{{{f}_{obs}}}}}{\Delta {{T}_{{{f}_{cal}}}}}=\dfrac{0.24{}^\circ C}{0.124{}^\circ C}=1.93$
Now, we will calculate the degree of dissociation of potassium chloride using this value of van't' hoff’s factor.
$\alpha =\dfrac{i-1}{n-1}$ where $\alpha $ is the degree of dissociation of the potassium chloride and $n$ is the number of ions, which is $2$ in case of $KCl$ for ${{K}^{+}}$and $C{{l}^{-}}$
So, substituting all these values in the equation we get,
$=\dfrac{1.93-1}{2-1}=0.93$
So the value of $\alpha $ came out to be $0.93$.

Note:The value of degree of dissociation can be calculated by using the freezing point of a solution. At first, we calculate the freezing point of the solution by using the given mass of the potassium chloride, molecular mass of the same and the weight of the solvent which was water in this case.
Then we write the observed freezing point of the solution and take the ratio of the two in order to calculate the value of van't’ Hoff’s factor. Now using this value we calculate the degree of dissociation by taking the number of ions to be two, as the potassium chloride would dissociate into potassium and chloride ions.