91Ó°ÊÓ

At standard atmospheric pressure, the freezing point may be defined as the temperature at which the liquid is converted into a solid.

\({\bf{\Delta Tb = Kb \times m}}\)

Determine the boiling point elevation and the boiling point.

\(\begin{aligned}{\rm{\Delta }}{{\rm{T}}_{\rm{f}}}\; &= \;\left( {{\rm{0}}{\rm{.}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C - 0}}{\rm{.9}}{{\rm{3}}^{\rm{^\circ }}}\;{\rm{C}}} \right)\\{\rm{\Delta }}{{\rm{T}}_{\rm{f}}} &= \;{\rm{0}}{\rm{.93^\circ C}}\end{aligned}\)

\(\begin{aligned}{{\rm{K}}_{\rm{f}}}{\rm{ \times m }} &= {\rm{1}}{\rm{.86^\circ C}}{{\rm{m}}^{{\rm{ - 1}}}}{\rm{ \times m(NaCl)}}\\{\rm{m(NaCl)}} &= \;\dfrac{{{\rm{0}}{\rm{.93^\circ C}}}}{{{\rm{1}}{\rm{.86^\circ C}}}}\end{aligned}\)

\(\begin{aligned}{\rm{\Delta }}{{\rm{T}}_{\rm{b}}} &= \;{{\rm{K}}_{\rm{b}}}{\rm{ \times m}}\\{\rm{\Delta }}{{\rm{T}}_{\rm{b}}} &= \;{\rm{0}}{\rm{.512^\circ C}}{{\rm{m}}^{{\rm{ - 1}}}}{\rm{ \times 0}}{\rm{.50}}\;{\rm{m}}\\{\rm{\Delta }}{{\rm{T}}_{\rm{b}}} &= \;{\rm{0}}{\rm{.256^\circ C}}\end{aligned}\)

\({\rm{\Delta }}{{\rm{T}}_{\rm{b}}}\;{\rm{ = }}\;{{\rm{T}}_{\rm{b}}}\left( {{\rm{given}}\;{\rm{solution}}} \right){\rm{ - }}{{\rm{T}}_{\rm{b}}}\left( {{\rm{water}}} \right)\)

The boiling point of water = is 100.00 °C.

By addition, it will give us the boiling point of NaCl solution,

\({\rm{100}}{\rm{.00 ^\circ C + 0}}{\rm{.26 ^\circ C = 100}}{\rm{.26 ^\circ C}}\)