Question

What is the freezing point of a solution of dibromobenzene,\({{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{4}}}{\bf{B}}{{\bf{r}}_{\bf{2}}}\), in 0.250 kg of benzene, if the solution boils at 83.5 °C?

Short answer

The freezing point of a solution of dibromobenzene, \({{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{4}}}{\rm{B}}{{\rm{r}}_{\rm{2}}}\) = -1.4^oC

Step-by-step solution

Definition of freezing point

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

Explanation 

Using the given value, determine the freezing point depression and the freezing point.

\(\begin{aligned}{\rm{\Delta T}} &= \;{\rm{83}}{\rm{.5^\circ C - 80}}{\rm{.1^\circ C}}\\{\rm{\Delta T}} &= \;{\rm{3}}{\rm{.4^\circ C}}\end{aligned}\)

\(\begin{aligned}{\rm{\Delta T}} &= \;{{\rm{K}}_{\rm{b}}}{\rm{ \times m}}\\{\rm{\Delta T}} &= \;{\rm{2}}{\rm{.53^\circ C}}\;{{\rm{m}}^{{\rm{ - 1}}}}{\rm{ \times m}}\\{\rm{3}}{\rm{.4^\circ C}} &= \;{\rm{2}}{\rm{.53^\circ C}}\;{{\rm{m}}^{{\rm{ - 1}}}}{\rm{ \times m}}\\{\rm{m - dibromobenzene}} &= \;\dfrac{{{\rm{3}}{\rm{.4^\circ C}}}}{{{\rm{2}}{\rm{.53^\circ C}}\;{{\rm{m}}^{{\rm{ - 1}}}}{\rm{ \times m}}}}\\{\rm{m - dibromobenzene}} &= \;{\rm{1}}{\rm{.34}}\;{\rm{m}}\end{aligned}\)

\(\begin{aligned}{\rm{\Delta T}} &= \;{{\rm{K}}_{\rm{f}}}{\rm{m}}\\{\rm{\Delta T}} &= \;{\rm{5}}{\rm{.12^\circ C}}\;{\rm{m}}{}^{{\rm{ - 1}}}{\rm{ \times 1}}{\rm{.34}}\;{\rm{m}}\\{\rm{\Delta T}} &= \;{\rm{6}}{\rm{.86^\circ C}}\end{aligned}\)

The freezing point of pure benzene is 5.5 °C.

Subtraction gives,

\({\rm{5}}{\rm{.5 ^\circ C - 6}}{\rm{.86 ^\circ C = -1}}{\rm{.4 ^\circ C}}{\rm{.}}\)