What is the molarity of the standard solution if the solubility product for a salt of type AB is $4\times {10}^{-8}$?

1. $2\times {10}^{-4}$ $\mathrm{mol}/\mathrm{L}$

2. $16\times {10}^{-16}$ $\mathrm{mol}/\mathrm{L}$

3. $2\times {10}^{-16}$ $\mathrm{mol}/\mathrm{L}$

4. $4\times {10}^{-4}$ $\mathrm{mol}/\mathrm{L}$

The solubility of $\mathrm{Ni}{\left(\mathrm{OH}\right)}_2$ in 0.1 M NaOH is:

K_sp ($\mathrm{Ni}{\left(\mathrm{OH}\right)}_2$)= 2x ${10}^{-15}$

1. 2 x ${10}^{-8}$ M.

2. 1 x ${10}^{-13}$ M

3. 1 x ${10}^8$ M 

4. 2 x ${10}^{-13}$ 

At room temperature, MY and NY₃, two nearly insoluble salts, have the same K_sp values of 6.2 × 10^-13. The true statement regarding MY and NY₃ is:

The minimum volume of water required to dissolve 1g of calcium sulphate at 298 K is

(For CaSO₄ , K_sp is 9.1 × 10^–6)

1. 1.22 L

2. 0.69 L

3. 2.44 L

4. 1.87 L

The solubility of BaSO₄ in water is $2.42$ $\times$ ${10}^{-3}$ g/ litre at 298 K. The value of the solubility product will be: (Molar mass of BaSO₄ = 233 gmol^–1)

The K_sp of Ag₂CrO₄ and AgBr is 1.1 × 10^–12 and 5.0 × 10^–13 respectively.