Which one of the following aqueous solutions will exhibit highest elivation in boiling point?

Question

Which one of the following aqueous solutions will exhibit highest elivation in boiling point? (A) 0.05 M glucose (non ionisable) (B) 0.01 M KNO3 (50 % ionisable) (C) 0.015 M Urea (non ionisable) (D) 0.01 M Na2SO4 (75 % ionisable)

Tardigrade Admin · Accepted Answer

Boiling point =T0(s o l v e n t)+Δ Tb(e l e v a t i o n i n b . p .) Δ Tb=miKb where, m is the molality (≈ m o l a r i t y) If n is number of species formed after dissociation i, the vant Hoff factor =[1 + (n - 1) α ] Kb= molal elevation constant. Thus Δ Tb propto i Assume 100% ionization [1]m i (N a2 S O4)=0.01[1 + (3 - 1) 0.75]=0.025 [2]m i (K N O3)=0.01× [1 + (2 - 1) 0.50]=0.015 [3]m i (u r e a)=0.015 [4]m i (g l u c o s e)=0.05

Q. Which one of the following aqueous solutions will exhibit highest elivation in boiling point?

0.05 M glucose (non ionisable)

0.01 M $K N O_{3}$ (50 % ionisable)

0.015 M Urea (non ionisable)

0.01 M $N a_{2} S O_{4}$ (75 % ionisable)

Q. Which one of the following aqueous solutions will exhibit highest elivation in boiling point?

0.05 M glucose (non ionisable)

0.01 M KNO3​ (50 % ionisable)

0.015 M Urea (non ionisable)

0.01 M Na2​SO4​ (75 % ionisable)

0.01 M $K N O_{3}$ (50 % ionisable)

0.01 M $N a_{2} S O_{4}$ (75 % ionisable)