A) \[1.87\times {{10}^{5}}\]
B) \[1.82\times {{10}^{4}}\]
C) \[6.24\times {{10}^{4}}\]
D) \[6.24\times {{10}^{5}}\]
Correct Answer: A
Solution :
| given, \[{{\operatorname{K}}_{4}}\left[ Fe{{\left( CN \right)}_{6}} \right]=\frac{1}{10}N\] \[\therefore {{\operatorname{K}}_{4}}\left[ Fe{{\left( CN \right)}_{6}} \right]=\frac{1}{10}\times \frac{1}{4}\operatorname{M}\] \[\left( N=M\times valence\,factor \right)\]\[\therefore For\,{{K}_{4}}\left[ Fe{{\left( CN \right)}_{6}} \right],n/V=\frac{1}{40}mol\,litr{{e}^{-1}}\]\[=\frac{1}{40}\times {{10}^{3}}\operatorname{mol}\,{{m}^{-3}}=25\,\operatorname{mol}\,{{m}^{-3}}\] |
| \[\because {{\pi }_{\operatorname{N}}}\times \operatorname{V}=\operatorname{nRT}\Rightarrow {{\pi }_{\operatorname{N}}}=\frac{n}{V}RT\] \[\therefore {{\pi }_{\operatorname{N}}}=25\times 8.314\times 300=6.236\times {{10}^{4}}{{\operatorname{Nm}}^{-2}}\] |
| For dissociation of \[{{\operatorname{K}}_{4}}Fe{{\left( CN \right)}_{6}}\]\[\begin{align} & {{\operatorname{K}}_{4}}\left[ Fe{{\left( CN \right)}_{6}} \right]4{{K}^{+}}+{{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}} \\ & \underset{\underset{{}}{\mathop{1\,-\,\alpha }}\,}{\mathop{1}}\,\,\,\,\,\,\,\,\,\,\,\,\underset{\underset{{}}{\mathop{4\alpha }}\,}{\mathop{0}}\,\underset{\alpha }{\mathop{0}}\, \\ \end{align}\] |
| Where \[\alpha \]is degree of dissociation |
| Given \[\alpha =0.5\] |
| Now \[{{\pi }_{\exp }}/{{\pi }_{N}}=1+4\alpha \] |
| Or \[{{\pi }_{\exp }}={{\pi }_{N}}\left( 1+4\alpha \right)\] |
| \[=6.236\times {{10}^{4}}\left( 1+4\times 0.5 \right)\] \[=1.87\times {{10}^{5}}{{\operatorname{Nm}}^{-2}}\] |
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