A) \[12\times {{10}^{-4}}M\]
B) \[12\times {{10}^{-3}}M\]
C) \[12\times {{10}^{-2}}M\]
D) \[12\times {{10}^{-1}}M\]
Correct Answer: C
Solution :
[c] At eqm. \[\begin{align} & 2HX(g)\rightleftharpoons \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{H}_{2}}(g)\,\,\,\,\,+\,\,\,\,\,{{X}_{2}}(g) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1.2\times {{10}^{-3}}M\,\,\,\,\,\,\,\,1.2\times {{10}^{-4}}M \\ \end{align}\] \[{{K}_{eq}}=\frac{[{{H}_{2}}][{{X}_{2}}]}{{{[HX]}^{2}}}\] \[{{10}^{-5}}=\frac{1.2\times {{10}^{-3}}\times 1.2\times {{10}^{-4}}}{{{[HX]}^{2}}}\] \[[HX]=\sqrt{\frac{1.2\times 1.2\times {{10}^{-7}}}{{{10}^{-5}}}}\] \[=1.2\times {{10}^{-1}}\] \[=12\times {{10}^{-2}}M\]
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