Problem Set 11 Solutions

1410. A solution containing 1.470 g of an unknown compound in 50.000 g of benzene boils at 80.60ºC at a totalpressure of 1 bar. The boiling point of pure benzene is 80.09ºC and the molar enthalpy of vaporization ofbenzene is 32.0 kJ/mol. Determine the molar mass of the unknown compound.In order to determine the molality of the unknown compound, we must first calculate the boiling point elevationconstant K b of the solvent, benzene, since it was not given in the problem. The boiling point elevationconstant K b is defined as€€K b =( ) 2*M solvent R T vap,( 1000 g/kg) ΔH vap,mwhere M solvent is the molar mass of the solvent (in this case, benzene), is the boiling point of the puresolvent, ΔH vap,m is the enthalpy € of vaporization of the solvent, and the factor of 1000 carries units of g/kg toconvert mass units.€€€ Substituting the values for benzene yields the boiling point elevation constant,K b ==( ) 2*M solvent R T vap( 1000 g/kg) ΔH vap,m( 78.69 g/mol) 8.314 J/molKK b = 2.55 K/molal .€Now that the boiling point elevation constant has been determined, the molality of the unknown compound maybe calculated. The equation for boiling point elevation is*T vap( )( 353.24 K) 2( 1000 g/kg) 32.0 ×10 3 J/mol( )Note that here, the unit 'm' indicates molality in moles of solute per kg of solvent.ΔT b = K b m solute .Solving for the molality of the solute and substituting gives€€m solute =m solute = ΔT bK b353.75 K − 353.24 K=2.55 K/mm solute = 0.200 molal.The moles of the unknown compound may be calculated from the molality,ormol solutekg solventmol solute = m solute ⋅ kg solvent,( )( 0.0500 kg solvent)= 0.200 mol solute/kg solventmol solute = 0.0100 moles.€