A sample of moonshine claiming to be 100 alcohol is submitt

A sample of “moonshine” claiming to be 100 % alcohol is submitted to you for analysis. 1.00 gram of “moonshine” is mixed with 1.50 grams of water. The normal boiling point of water is 100.0°C and kb for water is 0.51 °C/m. If the moonshine is pure ethanol, C2H5OH, 46.08 g/mol, what will be the solution’s boiling point?

Solution

mass of moonshinge = 1.00 g

moles = 1.00 / 46.08 = 0.0217 mol

molality = moles / mass of solvent in kg

             = 0.0217 / 0.0015

molality = 14.5 m

delta Tb = Kb x m

Tb - 100 = 0.51 x 14.5

Tb = 107.38 oC

solution’s boiling point = 107.38 oC