Question

What is the boiling point of a solution of 9.04 g of ${\mathbf{I}}_{\mathbf{2}}$in 75.5 g of benzene, assuming the ${\mathbf{I}}_{\mathbf{2}}$is non-volatile?

1. Outline the steps necessary to answer the question.
2. Answer the question.

Short answer

b) The boiling point of the solution is 354.3 K.

Step-by-step solution

Boiling Point

Boiling point may be defined as the temperature at which liquid converted into vapours at standard atmospheric pressure.

Subpart (a)

1. Find the molality of the given solution: $\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{=}\frac{\mathbf{N}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{o}\mathbf{f}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}}{\mathbf{M}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{o}\mathbf{f}\mathbf{S}\mathbf{o}\mathbf{l}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{(}\mathbf{k}\mathbf{g}\mathbf{)}}$
2. This is important because of the equation we use to determine the boiling point is

$\mathbf{\Delta }\mathbf{T}\mathbf{b}\mathbf{=}\mathbf{i}\times \mathbf{K}\mathbf{b}\times \mathbf{m}$

where

$\mathrm{\Delta }{T}_b=T_{solution}-T_{solution}$

T is the boiling point

m= molality,

$\mathbf{K}\mathbf{b}$cryoscopic constant - depends on the solvent
m- themolality of the solution;
i - the Van't Hoff factor - the number of ions per individual molecule of solute.

Subpart (b)

The number of moles of iodine:

$\begin{array}{rl}{n}_{I_2}& =\frac{m_{I_2}}{M_{I_2}}\\ & =\frac{9.04g}{253.8g/mol}\\ & =0.0356mol\end{array}$

Molality of the solution:

$\begin{array}{rl}m& =\frac{0.0356mol}{0.0755kg}\\ & =0.472mol/kg\end{array}$

The boiling point of the solution:

$\begin{array}{rl}\mathrm{\Delta }Tb& =i\times Kb\times m\\ T-353.1K& =1\times 2.53Kkg.mo{l}^{-1}\times 0.472mol/kg\\ Onsimplification,\\ T& =354.3K\end{array}$