Question

What is the freezing point of a solution containing 6.5 % KCl by mass (in water)?

Answer 1

Hint: Freezing point of any solution is directly proportional to molality of that solution. The proportionality sign is replaced by ${{K}_{f}}$ which is a molal depression constant. The value of molal depression constant for KCl is $1.853\,K\,kg\,mo{{l}^{-1}}$. For an electrolyte, Van’t Hoff factor is taken into consideration while calculating the freezing point.
Formula used:
Formula for depression in freezing point,$\Delta {{T}_{f}}={{K}_{f}}.m$, where m is molality, ${{K}_{f}}$ is molal depression constant, and $\Delta {{T}_{f}}$, freezing point depression.

Complete answer:
We have been given a solution having 6.5 % KCl by mass in water that means water is the solvent. 6.5 % KCl by mass means 65 g of solute (KCl) is dissolved in water, making a solution of 1000 g. Now, we will take out the molality using this data. As, $molality=\dfrac{moles\,of\,solute}{kg\,of\,solvent}$ , we have $moles=\dfrac{given\,mass}{molar\,mass}$ , and kilograms of solvent as, (1000 – 65) g $\times {{10}^{3}}$ kg, so,
$molality=\dfrac{\dfrac{65\,g}{74.55g\,mo{{l}^{-1}}}}{(1000\,g-65g)\times {{10}^{-3}}kg\,{{g}^{-1}}}$,
So, molality = 0.933 mol/kg
Now, we have KCl, when added to water, dissociation takes place as it is an electrolyte, the reaction is,
$KCl(s)\xrightarrow{{{H}_{2}}O}{{K}^{+}}+C{{l}^{-}}$
Which means that Van’t Hoff factor, ‘i’ will be of 2, as the ions before are, 1 and after are 2, that is,
$i=\dfrac{2}{1}$ = 2
So, the depression in freezing point, having Van’t Hoff factor will be written as, $\Delta {{T}_{f}}=i.{{K}_{f}}.m$, substituting the respective given values, ${{K}_{f}}$given as $1.853\,K\,kg\,mo{{l}^{-1}}$, the freezing point will be,
$\Delta {{T}_{f}}=2\times 0.933\,mol\,k{{g}^{-1}}\times 1.853\,K\,kg\,mo{{l}^{-1}}$
$\Delta {{T}_{f}}=3.36\,K$
$\Delta {{T}_{f}}=(0.0-3.46){}^\circ C$
Freezing point is = $-3.46{}^\circ C$
Hence, the freezing point of the solution having 6.5% KCl in water is $-3.46{}^\circ C$.

Note:
The mass of solvent will be calculated as mass of solution – mass of solute, which is 1000 – 65 = 35 g. 1 gram of anything is converted into kilogram by conversion factor, 1 kg = 1000 g. so, 1g = ${{10}^{-3}}kg$. Van’t Hoff factor is calculated as, $i=\dfrac{no.\,of\,atoms\,after\,dissociation}{no.\,of\,atoms\,before\,dissociation}$ . so, here it is, $i=\dfrac{2}{1}$ = 2.