Question

0.1 molal-equeous solution of an electrolyte $\text{FeC}{\text{l}}_{\text{2}}$ is 80% ionised. The boiling point of the solution at 1 atm is:

$\left[{\text{K}}_{\text{6}}\left({\text{H}}_{\text{2}}\text{O}\right)=0.52\text{K}\text{kg}\text{mo}{\text{l}}^{-1}\right]$

• A. $273.17K$
• B. $273.19{}^{\circ }\text{C}$
• C. $373.19{}^{\circ }\text{C}$
• D. $100.17{}^{\circ }\text{C}$

Answer 1

Answer: (D)

$100.17{}^{\circ }\text{C}$

$\underset{1-0.8}{\text{FeC}{\text{l}}_2}\to \underset{0.8}{\text{F}{\text{e}}^{2+}}+\underset{2\times 0.8}{2\text{C}{\text{l}}^{-}}$
$\text{i}=1-0.8+0.8+\left(2\times 0.8\right)$
(Van't Hoff Factor)
= 1 + 1.6 = 2.6
$\Delta {\text{T}}_{\text{b}}={\text{T}}_{\text{b}}-\text{T}{}_{\text{b}}^{\circ }=\text{i}{\text{K}}_{\text{b}}\text{m}$
Where,
${\text{T}}_{\text{b}}$ = boiling point of solution
$\text{T}{}_{\text{b}}^{\circ }$ = boiling point of pure water
m = molality of solutio
${\text{K}}_{\text{b}}\left(\text{water}\right)={0.52}^{\circ }\text{C}\text{kg}\text{mo}{\text{l}}^{-1}$
m = 0.1
$\Delta {\text{T}}_{\text{b}}=2.6\times 0.52\times 0.1$
$\Delta {\text{T}}_{\text{b}}={0.1352}^{\circ }\text{C}$
${\text{T}}_{\text{b}}=\Delta {\text{T}}_b+{{\text{T}}_{\text{b}}}^{\circ }$
= 0.1352 + 100
${\text{T}}_{\text{b}}={100.14}^{\circ }\text{C}$
Thus option (D) is correct.