What is the freezing point of a solution containing 8.1 g HBr in 100 g of water. Assuming the acid to be 90% ionised? (Kf for water = 1.86 K kg mol-1)

∆Tf=i×Kf×m;         HBr1-α→ H+α+Br-αTotal ions =1-α+α+α=1+α   ∴                    i=1+αGiven,          Kf=1.86 K kg mol-1 Mass of HBr=8.1 g  Mass of H2O=100 g (α)=degree of ionisation =90%molality( m) =mass of solute/molecular weight of solutemass of solvent(in kg)m(molality)=8.1/81100/1000                    i=1+α=1+90100=1.9              ∆Tf=i×Kf×m                      =1.9×1.86×8.1×81100×1000=3.534oC  ∆Tf(depression in freezing point)    = freezing point of water - freezing point of solution    3.534 = 0 - freezing point of solution ∴Freezing point of solution = -3.534o C.