The solubility of Ca(OH)₂ in water is: 

[Given: K_sp Ca(OH)₂ in water = 5.5 × 10^–6]

1. 1.77 × 10^–6

2. 1.11 × 10^–6

3. 1.11 × 10^–2

4. 2.77 × 10^–2

At 1990 K and 1 atm pressure, there are an equal number of Cl₂ molecules and Cl atoms in the reaction mixture.
The value of K_P for the reaction ${\mathrm{Cl}}_{2\left(\mathrm{g}\right)}\rightleftharpoons 2{\mathrm{Cl}}_{\left(\mathrm{g}\right)}$ under the above conditions is x × 10^–1. The value of x is:

1. 4

2. 8

3. 5

4. 10

For the reaction A_(g)$\to$(B)_(g), the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of $∆$_rG for the reaction at 300 K and 1 atm in J mol^–1 is – xR, where x is:
(R = 8.31 J mol^–1 K^–1 and ln 10 = 2.3)

1. 1400

2. 1380

3. 1360

4. 1340

An aqueous solution contains 0.10 M H₂S and 0.20 M HCl. If the equilibrium constant for the formation of HS^– from H₂S is 1.0 × 10^–7 and that of S^2– from HS^– ions is 1.2 × 10^–13 then the concentration of S^2– ions in aqueous solution will be:

1. 5 × 10^–8 M

2. 3 × 10^–20 M

3. 6 × 10^–21 M

4. 5 × 10^–19 M

An aqueous solution contains an unknown concentration of Ba²⁺. When 50 mL of a 1 M solution of Na₂SO₄ is added, BaSO₄ just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO₄ is 1 × 10^–10. What is the original concentration of Ba²⁺ ?

1. 5 × 10^–9 M

2. 2 × 10^–9 M

3. 1 × 10^–9 M

4. 1.0 × 10^–10 M

The solubility product of Pbl₂ is 8.0 × 10^–9 . The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10^–6 mol/L. The value of x is:

(Rounded off to the nearest integer)$\text{\hspace{0.17em}[Given:\hspace{0.17em}}\sqrt{2}=1.41\text{\hspace{0.17em}]\hspace{0.17em}}$

Arrange the following solution in the decreasing order of pOH :

(A). 0.01 M HCl

(B). 0.01 M NaOH

(C). 0.01 M ${\mathrm{CH}}_3\mathrm{COONa}$

(D). 0.01 M NaCl

1. (B) > (C) > (D) > (A)

2. (A) > (C) > (D) > (B)

3. (B) > (D) > (C) > (A)

4. (A) > (D) > (C) > (B)

The correct expression for the following reaction is:

Fe₂N_(s) + \(\frac{3}{2}\)H_2(g) \(\leftrightharpoons \) 2Fe_(s) + NH_3(g)

The variation of the equilibrium constant with temperature is given below :

Temperature               Equilibrium Constant

${\mathrm{T}}_1=25°\mathrm{C}$                   ${\mathrm{K}}_1=100$

${\mathrm{T}}_2=100°\mathrm{C}$                 ${\mathrm{K}}_2=100$

The values of $∆\mathrm{H}°,<mspace\; width="1em"></mspace>∆\mathrm{G}°<mspace\; width="1em"></mspace>\mathrm{at}<mspace\; width="1em"></mspace>{\mathrm{T}}_1<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>∆\mathrm{G}°<mspace\; width="1em"></mspace>\mathrm{at}<mspace\; width="1em"></mspace>{\mathrm{T}}_{2<mspace\; width="1em"></mspace>}\left(\mathrm{in}<mspace\; width="1em"></mspace>\mathrm{kJ}<mspace\; width="1em"></mspace>{\mathrm{mol}}^{-1}\right)$ respectively are close to: [use R=8.314 ${\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$]

1. 28.4, -7.14 and -5.71

2. 0.64, -7.14 and -5.71

3. 0.64, -5.71 and -14.29

4. 28.4, -5.71 and -14.29

Given that the equilibrium constant (K_C) at 800 K for the reaction \(N_2(𝑔)+3H_2(𝑔)⇋2NH_3(𝑔)\) is 64. What is the equilibrium constant K_C at the same temperature for the reaction \(NH_3(g) ⇌ \dfrac{1}{2}N_2(g) + \dfrac{3}{2}H_2(g)\)?