What is the boiling point of a solution that contains 3 moles of KBr in 2000g of water? a) 97 Degrees C b) 99.7 Degrees C c)101.4 Degrees C d) 103 degrees c
When a nonvolatile esolute is dissolved in water, according to Raoult’s laws, there will be an elevatio in the boiling point of water. The extent of elevation is given by
`DeltaTb = i*Kb *m` , where i is the number of species obtained from the dissociation (or association)of one molecule of the solute in solution, Kb is a constant, called molal ebullioscopic constant for a particular solvent, and m is the molal strength of the solute.
For water, Kb = 0.512,
Molality is the number of moles of solute dissolved per 1Kg solvent.
Here, 3 moles of KBr is dissolved in 2Kg water, therefore molality = 3/2 = 1.5.
KBr is an electrolyte, completely dissociating in aqueous solution as
KBr (s)→ K+(aq.) + Br-(aq.)
Hence i = 2.
Putting these values in equation,
ΔTb = 2*0.512 *1.5
Therefore, boiling point of the solution would be 100+1.536 = 101.5 degrees (approximately).
Correct answer is therefore, option c)101.4 Degrees C.