Question

If Kb for water is 0.512°C/m, then what is the boiling point of a solution that contains 1.25 mole CaCl2 in 1400 g of water? Show your work.

Answer 1

The new boiling point is found essentially the same way as the freezing point, except replacing the freezing point quantities with boiling point quantities:

`\Delta T_b=K_bm`

Where,

DeltaTb, is the change in boiling point temperature

Kb, is the molal boiling point

m, is the molality

We have to calculate first which is the molality of the solution.

`\begin{gathered} \text{Molality}=\frac{\text{moles of solute}}{kg\text{ of disolvent}} \\ \text{Molality}=(1.25molCaCl_2)/(1.4kg)=0.89m \end{gathered}`

Now, we replace the know data into the first equation:

`\begin{gathered} \Delta T_b=K_bm \\ \Delta T_b=0.512(\degree C)/(m)*0.89m=0.46\degree C \end{gathered}`

The boiling point of a solution will be the boiling temperature of water alone plus the temperature delta found:

Tb=100°C + 0.46°C=100.46°C

The boiling point of the solution will be 100.46°C