# Question 3b3c3

Mar 9, 2017

${104.93}^{\circ} \text{C}$

#### Explanation:

The idea here is that the boiling point of the solution will actually be higher than the boiling point of pure water, which means that the first thing to do here will be to figure out the boiling-point elevation, i.e. the difference between the boiling point of the solution and that of pure water.

Your tool of choice here will be the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\Delta T = i \cdot {K}_{b} \cdot b}}}$

Here

• $\Delta T$ is the boiling-point elevation
• $i$ is the van't Hoff factor
• ${K}_{b}$ is the ebullioscopic constant of the solvent, which in your case is water
• $b$ is the molality of the solution

Water has an ebullioscopic constant equal to

${K}_{b} = {\text{0.512"""^@"C kg mol }}^{- 1}$

https://en.wikipedia.org/wiki/Ebullioscopic_constant

So, you know that your solution has a molality of

$b = {\text{9.63 m" = "9.63 mol kg}}^{- 1}$

You also know that the solute is not ionic, which means that it does not dissociate in aqueous solution to produce ions.

Consequently, the van't Hoff fact, which tells you the ratio that exists between the number of moles of solute dissolved to make the solution and the number of moles of particles of solute produced in solution, will be equal to $1$.

$i = 1 \to$ you dissolve one mole of solute, you get one mole of particles of solute in solution

Therefore, you can say that the boiling-point elevation will be equal to

DeltaT = 1 * "0.512" ""^@"C" color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("mol"^(-1)))) * 9.63 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#

$\Delta T = {4.93}^{\circ} \text{C}$

Since this represents the increase in the boiling point of the solution when compared with the boiling point of the pure solvent, ${T}_{b}^{\circ}$, you will have

${T}_{\text{b sol}} = {T}_{b}^{\circ} + \Delta T$

which gets you

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{T}_{\text{b sol" = 100^@"C" + 4.93^@"C" = 104.93^@"C}}}}}$

The answer is rounded to two decimal places.