91Ó°ÊÓ

At standard atmospheric pressure, the boiling may be defined as the temperature at which the liquid is converted into vapors.

The boiling point is:

\(\begin{aligned} {\bf{\Delta }}{{\bf{T}}_{\bf{b}}}{\bf{ = Kb \times m}}\\{\bf{\Delta }}{{\bf{T}}_{\bf{b}}}{\bf{ = Elevation in boiling point}}\\{\bf{Kb = Ebuloscopic constant}}\\{\bf{m = molality of solution}}\end{aligned}\)

From the elevation of boiling point, determine the molarity expected,

\(\begin{aligned}\begin{aligned}{\rm{Mole \;Cd}}{{\rm{I}}_{\rm{2}}}\; &= \dfrac{{{\rm{5}}{\rm{.6977g}}}}{{{\rm{366}}{\rm{.220gmol}}{{\rm{e}}^{{\rm{ - 1}}}}}}\\{\rm{Mole \; Cd}}{{\rm{I}}_{\rm{2}}}\; &= {\rm{0}}{\rm{.015558 mole}}\end{aligned}\\\begin{aligned} {\rm{Molality \; of \; Cd}}{{\rm{I}}_{\rm{2}}} &= \dfrac{{{\rm{0}}{\rm{.015558mol}}}}{{{\rm{0}}{\rm{.04469kg}}}}\\{\rm{Molality \; of \; Cd}}{{\rm{I}}_{\rm{2}}} &= {\rm{0}}{\rm{.3481moleK}}{{\rm{g}}^{{\rm{ - 1}}}}\end{aligned}\end{aligned}\)

The boiling point elevation expected if no ionization occurs is:

\(\begin{aligned}{\rm{\Delta Tb}} & = {\rm {Kb \times m }}\\{\rm{\Delta Tb }}\; &= {\rm{ 0}}{\rm{.512 ^\circ C }}{{\rm{m}}^{{\rm{ - 1}}}}{\rm{ \times 0}}{\rm{.3481m}}\\{\rm{\Delta Tb }}\;&= {\rm{0}}{\rm{.17}}{{\rm{8}}^{\rm{o}}}{\rm{C}}\end{aligned}\)

The actual boiling point increase is 0.181°C.

After the addition of \({\rm{CdI_2}}\)in water, it will increase the boiling point of water.

That means it releases the heat in the solution.