Question

The elevation of boiling point of 0.5 M K2SO4 solution is more than that of 0.5 M urea

solution. Why?​

Answer 1

Answer:

it is because of vantoff's factor.

Explanation:

elevation in boiling point is given by ,

ΔTb = i * Kb * m

i = vantoff factor

Kb = molal boiling-point elevation constant

m = molality

here , molality = no .of moles /wt. of solvent = $\frac{Molarity * 1}{1}$  = molarity

(since, no. of moles = molarity *  no. of liters of solvent) (let us take 1 liter of water which is equal to 1 kg of water so wt. of solvent =1)

so, in this case molality = molarity.

and Kb value is same for both the solutions and also given their molality ( =molarity) is also same .

but the difference occurs in vantoff factor (i)

for K2SO4 , i = 3 (as in solution 1 moles dissociates to give 3 moles of ions)

for urea , i =1 (As in solution it does not undergo dissociate to give ions so 1 mole remains as 1 mole)

as elevation in boiling point is a colligative property it depends on no. constituents( it is no. of ions or atoms or molecules ; mostly we consider no. of ions) of solute in solvent. and this is number is involved in this equation by using vantoff factor.

as i of K2SO4> i of urea → elevation in boiling point of K2SO4 > urea.