Question

Calculate the boiling point of a one molar aqueous solution of KBr( density = 1.06 g/ml ). $K_{b}$ for water is 0.52 kg/mol , atomic mass of K =39 and Br= 80.

Answer 1

Hint: The given problem can be solved by suing the boiling point formula as:
\[\Delta {{\rm{T}}_{\rm{b}}} = {\rm{i}}{{\rm{K}}_{\rm{b}}}{\rm{m}}\] \[......\left( 1 \right)\]
Where, i is Van’t hoff factor, \[{{\rm{K}}_{\rm{b}}}\] is molal elevation constant and m is molality of the solution.

Complete step by step answer:
Let us first determine the molality of the solution as:
Concentration of the solution = 1 M
Density of the solution = \[1.06\;{\rm{gm}}{{\rm{L}}^{ - 1}}\]
Molar mass of KBr = molar mass of K + molar mass of Br
= 39 + 80
= 119g/mol
Amount of the solute, KBr = 1 mol = 119 g
Volume of the solution = 1 L = 1000 mL
Mass of solution = density $\times$ volume
$\Rightarrow$ 1000 $\times$ 1.06
$\Rightarrow$ 1060g
Mass of solvent = mass of solution – mass of solute
$\Rightarrow$ 1060 - 119
$\Rightarrow$ 941g
Now, substitute the value in the molality formula, we get
\[{\rm{Molality}}\;{\rm{of}}\;{\rm{solution}} = \dfrac{{{\rm{number}}\;{\rm{of}}\;{\rm{moles}}\;{\rm{of}}\;{\rm{solute}}}}{{{\rm{mass}}\;{\rm{of}}\;{\rm{solvent}}\;{\rm{in}}\;{\rm{kg}}}}\]
$\Rightarrow$ \[ \dfrac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.941}}\;{\rm{kg}}}}\]
$\Rightarrow$ \[ 1.0626\;{\rm{m}}\]
Now, KBr dissociates as:
\[{\rm{KBr}} \to {{\rm{K}}^ + } + {\rm{B}}{{\rm{r}}^ - }\]
Number of particles after dissociation is 2. Therefore, Van’t Hoff factor (i) will be 2.
Substitute the values in equation (1), we get
\[\Delta {{\rm{T}}_{\rm{b}}} = {\rm{i}}{{\rm{K}}_{\rm{b}}}{\rm{m}}\]
$\Rightarrow$\[ 2 \times 0.52 \times 1.0626\]
$\Rightarrow$ \[= {\rm{1}}{\rm{.1051^\circ C}}\]
So, the boiling point of the solution \[ = 100 + 1.1051\]
$\Rightarrow$\[ 101.1051^\circ {\rm{C}}\]

Note:
In equation (1), the Van’t hoff Factor (i) is used when electrolyte is used. Potassium bromide (KBr) is an electrolyte dissociated to give potassium ion and bromide ion. Therefore, the value of Van’t hoff factor (i) comes out to be 2.