Question

Sea water is found to contain 5.85% NaCl and 9.50% MgCl2 by weight of the solution. Calculate it's normal boiling point. Assume 80% ionisation for NaCl and 50% ionisation of MgCl2 [Kb(H2O) = 0.51 K kg mol-1)].

Answer 1

We have 0.1 mol of NaCl and 0.1 mol of MgCl2 in around 85g water

On 80% dissociation, 1 mol of NaCl will produce 1.8 moles

And on 50% dissociation, 1 mol of MgCl2 will produce 2 moles

So, we have,

0.38 moles in 85g water

Molality = 4.5

∂T = 4.5 * 0.51

= 2.29

So, sea water will boil at 102.29 C