A) \[2N\]
B) \[4N\]
C) \[\frac{N}{2}\]
D) \[\frac{N}{4}\]
Correct Answer: B
Solution :
\[\because \]Basicity of\[{{H}_{2}}S{{O}_{4}}=2\] \[Normality=molarity\times basicity\text{ }of\text{ }acid\] \[=2\times 2=4\] \[\therefore \] \[2M\,{{H}_{2}}S{{O}_{4}}=4N{{H}_{2}}S{{O}_{4}}\]
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