MCQ
Charge required to liberate $11.5\,g $ sodium is
- ✓$0.5\,F$
- B$0.1\,F$
- C$1.5\,F$
- D$96500$ coulombs
Charge (in $F$) $=$ moles of $e^-$ used $=$ moles of $Na$ deposited
$ = \frac{{11.5}}{{23}}\;gm = 0.5\,Faraday$.
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$\mathop {Pt\,|\,C{l_2}\left( g \right)}\limits_{\left( {0.4\,bar} \right)} \,|\,\mathop {C{l^ - }\left( {aq.} \right)}\limits_{0.1\,M} \,||\,\,\mathop {C{l^ - }\left( {aq.} \right)}\limits_{0.01} \,|\,\mathop {C{l_2}\left( g \right)\,|\,Pt}\limits_{0.2\,bar} $
the measured potential at $298\, K$ is .............. $\mathrm{V}$
Assume $100 \%$ ionization for both $NaCl$ and $MgCl _2$
Given : $K _{ b }\left( H _2 O \right)=0.52\,K\,kg\,mol ^{-1}$
Molar mass of $NaCl$ and $MgCl _2$ is 58.5 and $95\,g$ $mol ^{-1}$ respectively.
$\begin{align}
\begin{matrix}
C{{H}_{3}}-CH-C{{O}_{2}}H \\
|\,\,\,\,\,\,\,\, \\
\underset{\oplus }{\mathop{N}}{{H}_{3}} \\
\end{matrix}\,\,\,\,\,\,p{{K}_{a}}=2.2 \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p{{K}_{b}}=4.4 \\
\end{align}$
$(i)\,\,(CH_3)_2CH-CH_2Br \xrightarrow{{{C_2}{H_5}OH}} (CH_3)_2CH-CH_2OC_2H_5 + HBr$
$(ii)\,\,(CH_3)_2CH-CH_2-Br \xrightarrow{{{C_2}{H_5}{O^- }}} (CH_3)_2CH-CH_2OC_2H_5 + Br^-$
The mechanisms of reaction $(i)$ and $(ii)$ are respectively
$[Co(en)_3]\,[Cr(C_2O_4)_3]$
$[Cu(NH_3)_4]\, [CuCl_4]$
$[Ni(en)_3]\,[Co(NO_2)_6]$