Question

Calculate the normal boiling point of a solution consisting of 1.0 gram of naphthalene dissolved in 100.0 grams of toluene. The normal boiling point of toluene is 110.7⁰C and the boiling point constant is 36.1 K

Answer 1

Lets calculate molality first

Molar mass of C10H8,

MM = 10*MM(C) + 8*MM(H)

= 10*12.01 + 8*1.008

= 128.164 g/mol

mass(C10H8)= 1.0 g

use:

number of mol of C10H8,

n = mass of C10H8/molar mass of C10H8

=(1 g)/(1.282*10^2 g/mol)

= 7.803*10^-3 mol

m(solvent)= 100 g

= 0.1 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(7.803*10^-3 mol)/(0.1 Kg)

= 7.803*10^-2 molal

lets now calculate ΔTb

ΔTb = Kb*m

= 36.1*0.078

= 2.8167 oC

This is increase in boiling point

boiling point of pure liquid = 110.7 oC

So, new boiling point = 110.7 + 2.8167

= 113.5167 oC

Answer: 113.52 oC