AgCl ⇌ Ag⁺ + Cl^–
   s           s     s + 0.1

Concentration of Cl^– is (s + 0.1) mol L^–1 because s mol L^–1 from ionization of AgCl and 0.1 mol L^–1 from ionization of 0.1 M NaCl.

Now, K_sp = [Ag⁺][Cl^– ]

$$ \Rightarrow $$ 1.6 × 10^–10 = s (s + 0.1)

$$ \Rightarrow $$ 1.6 × 10^–10 = s (0.1) {$$ \because $$ s << 0.1}

$$ \Rightarrow $$ s = 1.6 × 10^–9 M

PF₃ is lewis base because on P atom there is lone pair.

C₅H₅N + H₂O → C₅H₅N⁺H + OH^–

So, the amount of pyridine that forms pyridinium ion is $$\alpha $$.

Now, $$\alpha $$ = $$\sqrt {{{{K_b}} \over {conc.\,of\,pyridine}}} $$

= $$\sqrt {{{1.7 \times {{10}^9}} \over {0.10}}} $$

= 1.30 $$ \times $$ 10^-4

So, percentage of pyridine that forms pyridinium ion

= 1.30 × 10^–4 × 100

= 1.30 × 10^–2 = 0.013%

This is Clausius-Clapeyron equation.

$${{d\ln P} \over {dT}} = {{\Delta {H_v}} \over {R{T^2}}}$$