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

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 = ,         = 2.53 C/m

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

where,    m = molality =

                   =

                   = 1

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

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

                        =

                        = 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.

i believe the answer is true.