Question

Calculate the percentage of pyridine (c5h5n) that forms pyridinium ion(c5h5nh+) 0.10m aquous pyridine solution

Answer 1

C5H5N + H2O ----> C5H5N^+ H + OH-

We know that,

α = √(Kb/M)

where

α --- degree of dissociation or association

Kb --- boiling point elevation constant

M --- Molarity

α = √(1.7×10^-9)/0.1

= √1.7×10^-8

= 1.3 × 10^-4

Percentage :

α % = 1.3 × 10^-4 × 100

= 0.013%

Hope it helps..!

Answer 2

Answer:

plzz refer the attachment below