Question

Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb.benzene boiling point=80.1 Kb=2.53carbon tetrachloride boiling point=76.8 Kb=5.03

Answer 1

Explanation:

We know that relation between change in temperature and molality is as follows.

           $\Delta T = m \times k_{b}$

And, boiling point of a solution is calculated as follows.

 Boiling point of a solution = Boiling point + $\Delta T (k_{c} \times m)$

For benzene solution, the boiling point is 80.1 + 2.53m.

For $CCl_{4}$ solution, the boiling point is 76.8 + 5.03m

Now, equating the boiling point of both the solutions as follows.

          80.1 + 2.53m = 76.8 + 5.03m    

                3.3 = 2.5 m

               m = 1.32 mol of solute per kg of solvent

Therefore, the boiling point will be calculated as follows.

             80.1 + 2.53m

         $80.1 + 2.53 \times 1.32$      

          = $83.43^{o}C$

Thus, we can conclude that molal concentration (m) is 1.32 and boiling point is $83.43^{o}C$.