Question

What is the boiling point of a 2.50 M solution of C6H4Cl2 in CCl? Assume C6H4Cl2 is not volatile

Answer 1

Solution :-

Normal boiling point of CCl4 = 76.8 C

Boiling point elevation constant of CCl4 = 5.02 m/C

Concentration of the C6H4Cl2= 2.50 M

So we have 2.50 mol of C6H4Cl2 in the 1 L solution

Using the density of the CCl4 we can find the mass of CCl4

Mass = volume x density

Mass of CCl4 = 1000 ml * 1.59 g/cm3

                         = 1590 g

1.590 g * 1 kg / 1000 g = 1.59 kg

Molality = moles / kg solvent

              = 2.50 mol / 1.59 kg

              = 1.57 m

Lets calculate the change in the boiling point of solution

Telta Tb= kb* m

              = 5.02 C/m * 1.57 m

             = 7.9 C

Boiling point of solution = boiling point of pure solvent + delta Tb

                                            = 77.8 C + 7.9 C

                                           = 85.7 C

Therefore the boiling point of solution will be 85.7 C