Mass of silver left = $ 0.162 \,gm$
Basicity of acid = $2$
Step $1 -$ To calculate the equivalent mass of the silver salt $(E)$
$\frac{{{\rm{Eq}}{\rm{.}}\,\,{\rm{mass}}\,\,{\rm{of}}\,\,{\rm{silver}}\,\,{\rm{salt}}}}{{{\rm{Eq}}{\rm{.}}\,\,{\rm{mass}}\,\,{\rm{of}}\,\,{\rm{silver}}}} = \frac{{{\rm{Mass}}\,\,{\rm{of}}\,\,{\rm{Acid}}\,\,{\rm{taken}}}}{{{\rm{Mass}}\,\,{\rm{of}}\,\,{\rm{silver}}\,\,{\rm{left}}}}$
$ \frac{E}{{108}} = \frac{{0.228}}{{0.162}}$
$ E = \frac{{0.228}}{{0.162}} \times 108 = 152\,({\rm{Eq}}{\rm{.}}\,\,{\rm{mass}}\,\,{\rm{of}}\,\,{\rm{silver}}\,\,{\rm{salt)}}$
Step $2$ -To calculate the eq. mass of acid.
Eq. mass of acid = Eq. mass of silver salt -Eq. mass of $Ag$ + Basicity
= $152 -108 + 1 = 152 -109 = 43$ (Eq. mass of acid)
Step $3 -$ To determine the molecular mass of acid.
Mol. mass of the acid = Eq. mass of acid $\times$ basicity = $45 \times 2 = 90$
$298\,K$ પર જ્યારે $\frac{\left[M^*(a q)\right]}{\left[M^{3 *}(a q)\right]}=10^a$ હોય ત્યારે આપેલ કોષ નો $E_{\text {cell }}$ એ $0.1115\,V$ છે. $a$ નું મૂલ્ય $............$ છે.આપેલ : $E _{ M }^\theta{ }^{3+} M ^{+}=0.2\,V$
$\frac{2.303\,R T}{F}=0.059\,V$
$Cr_{(s)} | Cr^{3+}_{(0.1\,M)} | | Fe^{2+}_{(0.01\,M)} | Fe;$
$E^0_{{cr}^{3+} |Cr} = -0.72 \,V,$ $ E^{0}_{{Fe}^{2+}{| Fe}}$ $= -0.42 \,V$
$Ag$ , $Ni$ , $Cr$
$Fe \rightarrow Fe^{2+} + 2e^{-} , E^{o} = 0.44\,\, V , 2H^{+} + 2e^{-} + \frac{1}{2} O_2 \rightarrow H_2O_{(l)}, E_{o} = 1.23\, V$ તો આ પ્રક્રિયા માટે $\Delta G^{o} =....$ કિલોજૂલ / મોલ