3.0 molal aqueous solution of an electrolyte A2B3 is 50 % ionised. The boiling point of the solution at 1 atm is: [Kb (H2 O) = 0.52 K k g m o l- 1]

Question

3.0 molal aqueous solution of an electrolyte A2B3 is 50 % ionised. The boiling point of the solution at 1 atm is: [Kb (H2 O) = 0.52 K k g m o l- 1] (A) 274.76 K (B) 377.68 K (C) 374.68 K (D) 104.68 K

Tardigrade Admin · Accepted Answer

A2B3(.aq.) arrow 2(A3)+(.aq.)+3B2 -(.aq.) n=5 Δ Tb=i.Kb.m =[.1+(.n-1.)α ].Kb.m. =[.1+4α ].Kb.m. =[.1+4(.0.5.)].0.52× 3 =(.1+2.)(.0.52.)× 3 =4.68 Ab=373+4.68=377.68K

Q. 3.0 molal aqueous solution of an electrolyte $A_{2} B_{3}$ is 50 % ionised. The boiling point of the solution at $1$ $a t m$ is: $[K_{b} (H_{2} O) = 0.52 K k g m o l^{- 1}]$

274.76 K

377.68 K

374.68 K

104.68 K

$A_{2} B_{3} (a q) \to 2 (A^{3})^{+} (a q) + 3 B^{2 -} (a q)$

$n = 5$

$Δ T_{b} = i . K_{b} . m$

$= [1 + (n - 1) α] K_{b} . m .$

$= [1 + 4 α] K_{b} . m .$

$= [1 + 4 (0.5)] 0.52 \times 3$

$= (1 + 2) (0.52) \times 3$

$= 4.68$

$A_{b} = 373 + 4.68 = 377.68 K$

Q. 3.0 molal aqueous solution of an electrolyte A2​B3​ is 50 % ionised. The boiling point of the solution at 1 atm is: [Kb​(H2​O)=0.52Kkgmol−1]