Suppose, solubility of PbCl$$_2$$ in water is s mol L$$^{-1}$$

$\begin{aligned} \mathrm{PbCl}_2(\mathrm{~s}) & \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+\underset{2 \mathrm{~s}}{2 \mathrm{Cl}^{-}}(\mathrm{aq}) \\ \mathrm{s}_{\mathrm{sp}} & =\left[\mathrm{Pb}^{2 \uparrow}\right] \cdot\left[\mathrm{Cl}^{-}\right]^2 \\ K_{\mathrm{sp}} & =[\mathrm{s}][2 \mathrm{~s}]^2=4 s^3 \\ 3.2 \times 10^{-8} & =4 \mathrm{~s}^3 \\ s^3 & =\frac{3.2 \times 10^{-8}}{4}=0.8 \times 10^{-8} \\ s^3 & =8.0 \times 10^{-9}\end{aligned}$

$$\begin{aligned} \text{Solubility of}\quad \mathrm{PbCl}_2, \mathrm{~s} & =2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \\ \text{Solubility of}\quad \mathrm{PbCl}_2 \text { in } \mathrm{gL}^{-1} & =278 \times 2 \times 10^{-3}=0.556 \mathrm{~g} \mathrm{~L}^{-1} \end{aligned}$$

$\left(\because\right.$ Molar mass of $\left.\mathrm{PbCl}_2=207+(2 \times 35.5)=278\right)$

0.556 g of $\mathrm{PbCl}_2$ dissolve in 1 L of water.

$\therefore \quad 0.1 \mathrm{~g}$ of $\mathrm{PbCl}_2$ will dissolve in $=\frac{1 \times 0.1}{0.556} \mathrm{~L}$ of water $$=0.1798 \mathrm{~L}$$

To make a saturated solution, dissolution of $0.1 \mathrm{g~PbCl}_2$ in $0.1798 \mathrm{~L} \approx 0.2 \mathrm{~L}$ of water will be required.

Although $\mathrm{BF}_3$ does not have a proton but acts as Lewis acid as it is an electron deficient compound. It reacts with $\mathrm{NH}_3$ by accepting the lone pair of electrons from $\mathrm{NH}_3$ and complete its octet. The reaction can be represented by

$$\mathrm{BF}_3+: \mathrm{NH}_3 \longrightarrow \mathrm{BF}_3 \leftarrow \mathrm{NH}_3$$

Lewis electronic theory of acids and bases can explain it. Boron in $\mathrm{BF}_3$ is $s p^2$ hybridised where N in $\mathrm{NH}_3$ is $s p^3$ hybridised.

Given that,

$\begin{aligned} \Delta_f H^{\ominus}[\mathrm{CaO}(s)] & =-635.1 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta_f H^{\ominus}\left[\mathrm{CO}_2(g)\right] & =-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta_f H^{\ominus}\left[\mathrm{CaCO}_3(s)\right] & =-1206.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$

In the reaction,

$\begin{gathered}\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(g) \\ \Delta_f H^{\ominus}=\Delta_f H^{\ominus}[\mathrm{CaO}(\mathrm{s})]+\Delta_f H^{\ominus}\left[\mathrm{CO}_2(g)\right]-\Delta_f H^{\ominus}\left[\mathrm{CaCO}_3(s)\right] \\ \therefore\quad\Delta_f H^{\ominus}=-635.1+(-393.5)-(-1206.9)=178.3 \mathrm{kJmol}^{-1}\end{gathered}$

Because $\Delta H$ value is positive, so the reaction is endothermic. Hence, according to Le-Chatelier's principle, reaction will proceed in forward direction on increasing temperature. Thus, the value of equilibrium constant for the reaction increases.

A. $\rightarrow(2)$

B. $\rightarrow(4)$

C. $\rightarrow(3)$

D. $\rightarrow$ (1)

A. Liquid $\rightleftharpoons$ Vapour equilibrium exists at the boiling point.

B. Solid $\rightleftharpoons$ Liquid equilibrium exists at the melting point.

C. Solid $\rightleftharpoons$ Vapour equilibrium exists at the sublimation point.

D. Solute $(s) \rightleftharpoons$ Solute (solution) equilibrium exists at saturated solution.

$$\mathrm{A.\to(4)\quad B.\to(3)\quad C.\to(2)}$$

For the reaction,

$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$

Equilibrium constant $K_{\mathrm{c}}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}$

A. The given reaction $\left[2 \mathrm{~N}_2(g)+6 \mathrm{H}_2(g) \rightleftharpoons 4 \mathrm{NH}_3(g)\right]$ is twice the above reaction. Hence, $K=K_c^2$

B. The reaction $\left[2 \mathrm{NH}_3(g) \rightleftharpoons \mathrm{N}_2(g)+3 \mathrm{H}_2(g)\right]$ is reverse of the above reaction. Hence, $K=\frac{1}{K_c}$

C. The reaction $\left[\frac{1}{2} \mathrm{~N}_2(g)+\frac{3}{2} \mathrm{H}_2(g) \rightleftharpoons \mathrm{NH}_3(g)\right]$ is half of the above reaction. Hence, $K=\sqrt{K_c}=K_c^{\frac{1}{2}}$.