A) \[\text{2N}\]
B) \[\text{4}\,\text{N}\]
C) \[\frac{N}{2}\]
D) \[\frac{N}{4}\]
Correct Answer: B
Solution :
\[\because \]Basicity of \[{{H}_{2}}S{{O}_{4}}=2\] Normality = molarity \[\times \]basicity of acid \[=2\times 2=4\] \[\therefore \] \[2M\,{{H}_{2}}S{{O}_{4}}=HN{{H}_{2}}S{{O}_{4}}\]You need to login to perform this action.
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