If you use 5.76mol of sodium fluoride (NaF) and dissolved this into 3.62 kg of water, what will be the change in the boiling point of your solution. Assume the Kb of water is 0.51°C/m.
∆Tb= iKbm
m= 5.76÷3.62= 1.5912
i= 2
∆Tb=2×0.51×1.5912
=1.62oc
The boiling point is elevated by 1.62oc
Comments
Leave a comment