A) \[91.3%\]
B) \[87%\]
C) \[100%\]
D) \[74%\]
Correct Answer: B
Solution :
\[\begin{matrix} {} \\ At\,t=0 \\ At\text{ }eqm \\ \end{matrix}\,\,\,\,\begin{matrix} Ba{{\left( N{{O}_{3}} \right)}_{2}} \\ 0.1\,M \\ (0.1-x)M \\ \end{matrix}\,\,\begin{matrix} \\ {} \\ {} \\ \end{matrix}\begin{matrix} B{{a}^{2+}} \\ 0 \\ xM \\ \end{matrix}\,\,\begin{matrix} + \\ {} \\ {} \\ \end{matrix}\,\,\,\begin{matrix} 2NO_{3}^{-} \\ 0 \\ 2x\,M \\ \end{matrix}\] \[i=\frac{(0.1-x)+x+2x}{0.1}\] \[2.74=\frac{0.1+2x}{0.1}\] \[0.1+2x=0.27\,4\] \[2x=0.274-0.1=0.174\] \[x=\frac{0.174}{2}=0.087\] \[\therefore \]degree of dissociation \[=\frac{0.087}{0.1}\times 100=87%\]
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