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 :
We will use expression\[K=\frac{[{{H}_{2}}][C{{l}_{2}}]}{{{[HCl]}^{2}}}\] Given,\[K=1.0\times {{10}^{-5}}[{{H}_{2}}]=1.2\times {{10}^{-3}}M\] \[K=\frac{[{{H}_{2}}][C{{l}_{2}}]}{{{[HCl]}^{2}}}\] or \[{{[HCl]}^{2}}=\frac{[{{H}_{2}}][C{{l}_{2}}]}{K}\] or \[[HCl]=\sqrt{\frac{[{{H}_{2}}][C{{l}_{2}}]}{K}}\] Substituting the values \[=\sqrt{\frac{1.2\times {{10}^{-3}}\times 1.2\times {{10}^{-4}}}{1.0\times {{10}^{-5}}}}\] \[=\sqrt{1.2\times 1.2\times {{10}^{-7}}\times {{10}^{5}}}\] \[=\sqrt{\frac{1.2\times 1.2}{100}}\] \[=0.12\text{ }M=12\times {{10}^{-2}}M\]
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