# Question 6a5d5

Jul 19, 2017

The raising of the boiling point is one of the colligative properties of solutions.

Being a colligative property, the effect is independent of the identity of the solute, but only the amount of solute in solution.

The equation for boiling point elevation is

$\Delta {T}_{b} = i m {K}_{b}$

where

• $\Delta {T}_{b}$ is the change in boiling point (a positive quantity) of the solution

• $i$ is the van't Hoff factor, which is essentially the number of dissolved ions per unit of solvent (assuming $\text{HCl}$ is the solute, this value is $2$, because there is $1$ ${\text{H}}^{+}$ ion and $1$ ${\text{Cl}}^{-}$ ion per unit of $\text{HCl}$)

• $m$ is the molality of the solution;

$\text{molality" = "mol solute"/"kg solvent}$

If we can figure out these two quantities, we can figure out the solution's molality.

• ${K}_{b}$ is the molal boiling point elevation constant for the solvent (assumed to be benzene). We can find this value from a list online:

From this, we can see that ${K}_{b}$ for benzene is color(red)(2.53 color(red)(""^"o""C/"m.

Plugging in the known values for the situation, we have

$\Delta {T}_{b} = \left(2\right) m \left(\textcolor{red}{2.53} \textcolor{w h i t e}{l} \textcolor{red}{\text{^"o""C/} m}\right)$

$\Delta {T}_{b} = \left(5.06 {\textcolor{w h i t e}{l}}^{\text{o""C/}} m\right) \cdot m$

As long as we know the molal concentration $m$, we can find the increase in boiling point of this solution.

Let's say there is $1.00$ $\text{mol HCl}$ dissolved in $2.00$ $\text{kg benzene}$.

The molality would be

"molality" = (1.00color(white)(l)"mol HCl")/(2.00color(white)(l)"kg benzene") = color(green)(0.500m

Plugging this into the equation, we can find the boiling point elevation:

$\Delta {T}_{b} = \left(5.06 {\textcolor{w h i t e}{l}}^{\text{o""C/}} \cancel{m}\right) \left(\textcolor{g r e e n}{0.500} \cancel{\textcolor{g r e e n}{m}}\right) = 2.53$ $\text{^"o""C}$

This quantity represents by how much the boiling point of the solvent (benzene) increases; to find the new boiling point of this solution, we simply add this to the normal boiling point of benzene (according to the above figure, it is $80.1$ $\text{^"o""C}$):

"new b.p." = 80.1^"o""C" + 2.53^"o""C" = color(blue)(82.6 color(blue)(""^"o""C"#

rounded to one decimal place.