The molar solubility of ${\mathrm{CaF}}_2$ $({\mathrm{K}}_{\mathrm{sp}}=5.3$ $\mathrm{\times }$ ${10}^{-11})$ in 0.1 M solution of NaF will be:

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

The solubility product of \(\mathrm{BaSO_4}\) in water is \(1.5 \times 10^{-9} \). The molar solubility of \(\mathrm{BaSO_4}\) in 0.1 M solution of Ba(NO₃)₂ in:
1. \(2.0 \times 10^{-8} M\)
2. \(0.5 \times 10^{-8} M\)
3. \(1.5 \times 10^{-8} M\)
4. \(1.0 \times 10^{-8} M\)

The solubility product for a salt of type AB is $4\times {10}^{-8}$.  The molarity of its standard solution will be:
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 molar solubility of Ni(OH)₂ in a solution containing 0.10 M NaOH, given that the ionic product of Ni(OH)₂ is 2.0 × 10^–15, is:

Given that the ionic product of $\mathrm{Ni}{\left(\mathrm{OH}\right)}_2$ is 2 × ${10}^{-\mathrm{15 }}$. The solubility of $\mathrm{Ni}{\left(\mathrm{OH}\right)}_2$ in 0.1 M NaOH is ;
1. 2 × ${10}^{-8}$ M

The values of K_sp of two sparingly soluble salts Ni(OH)₂ and AgCN are 2.0 × 10^–15 and 6 × 10^-17 respectively.
The more soluble salt is:

1. Ni(OH)₂ is more soluble than AgCN
2. AgCN is soluble in than Ni(OH)₂
3. Both have similar solubility
4. None of the above

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

When equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together (K_{sp  (cupric iodate)} = 7.4 × 10^–8 ), from the following, the correct observation would be :

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