JEE Main202124 Feb 2021Morning ShiftChemistrySolutionsActual
When 9 . 45 g of ClCH 2 COOH is added to 500 mL of water, its freezing point drops by 0 . 5 ° C . The dissociation constant of ClCH 2 COOH is x × 10 - 3 . The value of x is off to the nearest integer) K f H 2 O = 1 . 86 K kg mol - 1
Correct answer
0
Step-by-step solution
ClCH 2 COOH ⇌ ClCH 2 COO ⊖ + H + i = 1 + ( 2 - 1 ) α i = 1 + α ΔT f = i . K f . m 0 . 5 = ( 1 + α ) ( 1 . 86 ) 9 . 45 94 . 5 500 1000 5 3 . 72 = 1 + α ⇒ α = 1 . 28 3 . 72 α = 32 93 K a = C α 2 C − C α = C α 2 1 − α C = 0 .1 500 / 1000 = 0 .2 K a = 0 .2 32 / 93 2 1 − 32 / 93 = 0 .2 × 32 2 93 × 61 = 0 .036 K a = 36 × 10 − 3