What are the boiling point and freezing point of a 3.47 m solution of naphthalene in benzene? (the boiling point and freezing point of benzene are 80.1°c and 5.5°c respectively. the boiling point elevation constant for benzene is 2.53°c/m, and the freezing point depression constant for benzene is 5.12°c/

Elevation in boiling point is mathematically expressed as
ΔTb = Kb X m

where Kb =  boiling point elevation constant =  2.53°c/m (for benzene)
m = molality of solution = 3.47 m (given)

∴ ΔTb = 2.53 X 3.47
           = 8.779 oC

But, boiling point of benzene = 80.1 oC
∴ Boiling point of solution = 88.879 oC

Now, Depression in freezing point = ΔTf = Kf X m
where, Kf = freezing point depression constant = 5.12°c/m (for benzene)

∴ ΔTf = 5.12 X 3.47
         = 17.766 oC

But freezing point of benzene = 5.5 oC
∴ Freezing point of solution = -12.266 oC

Well to my expert knowledge and my exquisite calculations i think the the negative ion # 412 is the key to sucess