${\mathrm{PCl}}_5$ dissociates into ${\mathrm{PCl}}_3$ and ${\mathrm{Cl}}_2$. If the total pressure of the system in equilibrium is $\mathrm{p}$ at a density $\mathrm{d}$ and temperature $\mathrm{T},$ show that $\mathrm{\alpha }=\frac{\mathrm{pM}}{\mathrm{dRT}}-1$ where $\mathrm{\alpha }$ is the degree of dissociation, $\mathrm{M}$ is the relative molar mass of ${\mathrm{PCl}}_5$. If the vapour density has a value of $62$ when the temperature is $230°\mathrm{C}$, what is the value of $\mathrm{p}/\mathrm{d}?$

The decomposition of ${\mathrm{PH}}_3$ at $950 \mathrm{K}$ according to $4{\mathrm{PH}}_3\left(\mathrm{g}\right)\to {\mathrm{P}}_4\left(\mathrm{g}\right)+6{\mathrm{H}}_2\left(\mathrm{g}\right)$ is a first-order reaction. The following measurements were made on a system containing only ${\mathrm{PH}}_3$ initially.

Calculate the rate constant in ${\min }^{-1}$ .

The standard reduction potential of the electrode ${\mathrm{OH}}^{-}|{\mathrm{H}}_2 (\mathrm{Pt}) (1 \mathrm{atm})$ is ${\mathrm{E}}^0=0.828 \mathrm{V}$. Calculate ${\mathrm{K}}_{\mathrm{w}}$ at $298 \mathrm{K}$ if Nernst equation takes the form $\mathrm{E}={\mathrm{E}}_{{\mathrm{H}}_2\mathrm{O}, {\mathrm{OH}}^{-}}^0-0.059 \log \left[{\mathrm{OH}}^{-}\right]$. If ${\mathrm{K}}_{\mathrm{w}} =\mathrm{a}\times {10}^{-15}$, report $\text{'}\mathrm{a}\text{'}$ after rounding-off up to two places of decimals.

The standard reduction potential of the electrode ${\mathrm{Cl}}^{-}|\mathrm{CuCl}$ is ${\mathrm{E}}^0=0.137$ volt. Calculate ${\mathrm{K}}_{\mathrm{sp}}$ of $\mathrm{CuCl}$ at $298 \mathrm{K}. \left({\mathrm{E}}_{{\mathrm{Cu}}^{+},\mathrm{Cu}}^0=0.521 \mathrm{volt}\right).$ If your answer is $\mathrm{y} \times {10}^{-7}. \mathrm{then} \mathrm{y} \mathrm{is}$

Calculate the emf of the cell

$$\begin{gathered} \mathrm{Pb} \left| {\mathrm{PbSO}}_{4 } \right| \left| {\mathrm{PbI}}_{2 }\right| \mathrm{Pb} \\ \mathrm{sat}.\mathrm{soln}. \mathrm{sat}.\mathrm{soln}. \end{gathered}$$

at $25°\mathrm{C}$. ${\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{PbSO}}_4\right)=1.6\times {10}^{-8}$ and ${\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{PbI}}_2\right)=8\times {10}^{-9}$. 

If your answer is $\mathrm{y} \times {10}^{-3}$. then y is.