Question

A 175.3 mg sample of C10H12O2 was dissolved in 1.50 g of chloroform (Kb = 3.63 °C/m). What is the boiling point of this solution.

Answer 1

`Tb=63.8\°C`

Step-by-step explanation:

Hello,

In this case, for the boiling point elevation, we have:

`\Delta T=i*m*Kb`

Whereas the van't Hoff factor for the given substance is 1 since is not ionizing. Moreover, the molality is computed by:

`m=(mol_(solute))/(m_(solvent))=(175.3mg*(1g)/(1000mg)*(1nol)/(164g) )/(1.50g*(1kg)/(1000g) ) =0.713m`

In such a way, since the boiling point of pure chloroform is 61.2 °C, the boiling point of the solution is:

`Tb=61.2\°C+1*0.713m*3.63(\°C)/(m) \\\\Tb=63.8\°C`

Regards.