In an evacuated vessel of capacity $49.2 \mathrm{L},4$ moles of $Ar\left(\mathrm{g}\right)$ and $5$ moles of ${\mathrm{PCl}}_5\left(\mathrm{g}\right)$ were introduced and the temperature is maintained at $240 \mathrm{K}$ (assume that at this temperature decomposition of ${\mathrm{PCl}}_5$ takes place). At equilibrium, the total pressure of the mixture was found to be $4.8 \mathrm{atm}$. Correct statement(s) is/are

The following represent the variation of $\mathrm{pH}$ with acid strength $\mathrm{pKa}$ for $0.10\mathrm{M}$ aqueous solutions, Marked as $\mathrm{a},\mathrm{b}$ and $\mathrm{c}$.

Select the correct matchings:

At $1467°\mathrm{C}$, the progress of the reaction:
$\mathrm{xA}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{yB}\left(\mathrm{g}\right)$
in a $1\mathrm{L}$ vessel is presented in the following figure:

Identify the correct statement(s):

Determine the concentration of ${\mathrm{NH}}_3$ solution, one litre of which can dissolve $0.10 \mathrm{mole}$ $\mathrm{AgCl}. {\mathrm{K}}_{\mathrm{sp}}$ of $\mathrm{AgCl}$ and ${\mathrm{K}}_{\mathrm{f}}$ of $\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2^{+}$ are $1.0\times {10}^{-10} {\mathrm{M}}^2$ and $1.6\times {10}^7 {\mathrm{M}}^{-2}$ respectively.

How much of ${\mathrm{Ag}}^{+}$would remain in the solution after mixing equal volumes of $0.08 \mathrm{M}$ ${\mathrm{AgNO}}_3$ and $0.08 \mathrm{M} \mathrm{HOCN}$ ? ${\mathrm{K}}_{\mathrm{a} }\left(\mathrm{HOCN}\right)=3.3\times {10}^{-4} {\mathrm{K}}_{\mathrm{sp} }\left(\mathrm{AgOCN}\right)=2.3\times {10}^{-7}$

If the answer is of type $\mathrm{y}\times {10}^{-4} \mathrm{M}$, report the value of $\mathrm{y}$ correct up to nearest integer value.

Calculate $\left[{\mathrm{NH}}_4^{+}\right]$ (derived from ${\mathrm{NH}}_4\mathrm{Cl}$ ) to prevent $\mathrm{Mg}(\mathrm{OH}{)}_2$ from precipitating in a one-litre solution containing $0.01 \mathrm{M} {\mathrm{NH}}_3$ and $0.001 \mathrm{M} {\mathrm{Mg}}^{2+}$ ions.

${\mathrm{K}}_{\mathrm{sp}}\left[\mathrm{Mg}(\mathrm{OH}{)}_2\right]=1.2\times {10}^{-11}, {\mathrm{K}}_{\mathrm{b}}\left({\mathrm{NH}}_4\mathrm{OH}\right)=1.8\times {10}^{-5}$

[Answer has to be given after dividing with ${10}^{-3}$ and correct up to one place of decimal.]

Calculate the solubility product of the reaction

$\mathrm{Fe}(\mathrm{OH}{)}_3={\mathrm{Fe}}^{3+}+3{\mathrm{OH}}^{-}$ Given that, 

$\mathrm{Fe}\left(\mathrm{OH}{)}_3 \right(\mathrm{s})+3\mathrm{e}=\mathrm{Fe} (\mathrm{s})+3{\mathrm{OH}}^{-}; {\mathrm{E}}^0=-0.77 \mathrm{V}$

${\mathrm{Fe}}^{3+}+3\mathrm{e}=\mathrm{Fe} \left(\mathrm{s}\right); {\mathrm{E}}^0=-0.036 \mathrm{V}$