Question

What is the boiling point change for a soln containing 0.870 mol of naphthalene (a nonvolatile, nonionizing compound) in 250. g of liquid benzene? (Kb = 2.53°C/m for benzene)

Answer 1

Solution :-

moles of naphthalene = 0.870 mol

mass of benzene = 250 g

250 g * 1 kg / 1000 g = 0.250 kg

Lets calculate the molality of the solution

molality = moles / kg solvent

             = 0.870 mol / 0.250 kg

             = 3.48 m

formula to calculate the change in the boiling point is as follows

Delta Tb = Kb * m

             = 2.53 C/m * 3.48 m

             = 8.80 C

So the change in the boiling point is 8.80 C