We know,
$\mathrm{\pi }=\mathrm{iCRT}$

$\mathrm{\pi }\propto \mathrm{iC}$

A. $0.500 \mathrm{M} {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{aq}\right)$$\left(\mathrm{iC}=1\times 0.5=0.5\right)$ and $0.25 \mathrm{M} \mathrm{KBr}\left(\mathrm{aq}\right)\left(\mathrm{iC}=2\times 0.25=0.5\right)$

B. $0.100 {\mathrm{M}}_{} {\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]\left(\mathrm{aq}\right)\left(\mathrm{iC}=5\times 0.1=0.5\right)$ and $0.100 \mathrm{M}$ ${\mathrm{FeSO}}_4{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\left(\mathrm{iC}=5\times 0.1=0.5\right)$

C. $0.05 \mathrm{M} {\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]\left(\mathrm{iC}=5\times 0.05=0.25\right)$ (aq) and $0.25 \mathrm{M} \mathrm{NaCl}\left(\mathrm{iC}=2\times 0.25=0.5\right)$ $\left(\mathrm{aq}\right)$

D. $0.15 \mathrm{M} \mathrm{NaCl}\left(\mathrm{aq}\right)\left(\mathrm{iC}=2\times 0.15=0.3\right)$ and $0.1 \mathrm{M} {\mathrm{BaCl}}_2\left(\mathrm{aq}\right)\left(\mathrm{iC}=3\times 0.1=0.3\right)$

E. $0.02 \mathrm{M} \mathrm{KCl}.{\mathrm{MgCl}}_2.6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{aq}\right)\left(\mathrm{iC}=5\times 0.02=0.1\right)$ and $0.05 \mathrm{M}$ $\mathrm{KCl}\left(\mathrm{aq}\right)\left(\mathrm{iC}=2\times 0.05=0.1\right)$