Question

 

When a 2.5 mol of sugar (C₁₂H₂₂O₁₁) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. If 2.5 mol of aluminum nitrate is added to the same amount of water, by how much will the boiling point be changed? Show all calculations leading to your answer OR use 3 – 4 sentences to explain your answer.

Answer 1

Boiling point elevation, ∆T_b = K_b × b_solute × ¡

where,

K_b = ebullioscopic constant, for water, it is 0.512℃/m

b_solute = molality of solute

i = Van't Hoff factor

K_b is same for both sugar and ammonium nitrate solution because it depands on solvent ( water ) only

b_solute is same for both sugar and ammonium nitrate solution because amount of water and no of moles are same for both solutions

i is differing for sugar and ammonium nitrate

for Sugar , i = 1 because sugar is non electrolyte

for Ammonium nitrate , i = 2 because NH4NO3 dissociate completely into two ions i.e NH4+ and NO3-

So,

for sugar ,∆T_b = K_b × b_solute × 1

for Ammonium Nitrate, ∆T_b = K_b × b_solute ×3

Therefore,

if boiling point raise for Sugar solution is 1℃

boiling point raise for Ammonium nitrate solution is 2℃