NTA Abhyas JEE Main2020ChemistrySolutionsPractice
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
Options
- A274.76 K
- B377.68 K
- C374.68 K
- D104.68 K
Correct answer
B. 377.68 K
Step-by-step solution
A 2 B 3 ( a q ) → 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 × 3 = ( 1 + 2 ) ( 0.52 ) × 3 = 4.68 A b = 373 + 4.68 = 377.68 K