Question

assuming ideal behavior, calculate the boiling point in C of solution that contains 64.5g or the non-volatile non-electrolyte quinoline( molar mass= 129g/mole) in 500 grams of benzene. For benzene the normal boling point is 80.10 C and Kb = 2.53C/m

Answer 1

Explanation:

The given data is as follows.

     mass of solute = 64.5 g,        mass of solvent = 500 g = 0.5 kg (as 1 kg = 1000 g),  

    Temperature = $80.10^{o}C$,        $K_{b}$ = 2.53 C/m

Also, we know that the relation between change in temperature, van't Hoff factor(i) and $K_{b}$ is as follows.

         $\Delta T = iK_{b}m$

where,    m = molality = $\frac{\text{moles of solute}}{\text{kg of solvent}}$

                   = $\frac{64.5 g}{129 g/mol} \times \frac{1}{0.5 kg}$

                   = 1

It is known that for non-electrolyte solute i = 1.

Hence, putting the given values into the above formula as follows.

            $\Delta T = iK_{b}m$

                        = $1 \times 2.53 C/m \times 1$

                        = 2.53 C

Therefore, calculate the boiling point of the solution as follows.

                    80.10 C + 2.53 C

                  = 82.63 C

Thus, we can conclude that the boiling point of the solution is 82.63 C.