For the reaction, $3{\mathrm{O}}_2$ $\left(\mathrm{g}\right)$ $\rightleftharpoons$ $2{\mathrm{O}}_3$ $\left(\mathrm{g}\right),$ ${\mathrm{K}}_{\mathrm{c}}$ $=$ $2.0$ $\times$ ${10}^{-50}$ L/mol at 25^°C. If the equilibrium concentration of O₂ in air at 25^°C is $1.6$ $\times$ ${10}^{-2}$, the concentration of O₃ would be: 

$1.$ $2.86$ $\times {10}^{28}$ $\mathrm{M}$
$2.$ $28.66$ $\times$ ${10}^{-28}$ $\mathrm{M}$
$3.$ $1.43$ $\times$ ${10}^{-28}$ $\mathrm{M}$
$4.$ $2.86$ $\times$ ${10}^{-28}$ $\mathrm{M}$

Consider the following reaction:
\(CO_{\left(\right. g \left.\right)}\) \(+\) \(3 {H_{2}}_{\left(g\right)}\) \(\rightleftharpoons\) \({CH_{4}}_{\left(\right. g \left.\right)}\) \(\)\(+\) \(H_{2} O_{\left(\right. g \left.\right)}\) \(\)

If the reaction mixture contains 0.30 mol of CO, 0.10 mol of H_2, and 0.02 mol of H₂O, and an unknown amount of CH₄ at equilibrium at 1300 K in a 1L flask, the concentration of CH₄ in the mixture will be -

(K_c = 3.90 at 1300 K) $\mathrm{}$

1 . \(5 . 85\) \(\times\) \(10^2\) \(M\)
2 . \(5 . 85\) \(\times\) \(10^{- 1}\) \(M\)
3 . \(5 . 85\) \(\times\) \(10^3\) \(M\)
4 . \(5 . 85\) \(\times\) \(10^{- 2}\) \(M\)

The concentration of hydrogen ion in a sample of soft drink is $3.8$ $\times {10}^{-3}$ $\mathrm{M}$. The  pH of the soft drink will be:

What will be the concentration of H⁺ ions in a vinegar sample with a pH of 3.76?

$1.$ $8.95$
$2.$ $1.43$
$3.$ $3.56$
$4.$ $6.30$

(Atomic wt of Tl =204)

When 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution, the pH of the resultant solution will be:

A 0.001 M aniline solution has a pH of:
(\(K_b = 4.27 \times 10^{-10}\))

The degree of ionization of 0.05 M acetic acid is:

(pK_a = 4.74)

The percentage ionization of 0.02 M dimethylamine solution if it also contains 0.1 M NaOH solution (K_b of dimethylamine = 5.4 × 10^–4) will be: