A) 1 \[l-N-{{H}_{2}}S{{O}_{4}}\]
B) 1 \[l-M-{{H}_{2}}S{{O}_{4}}\]
C) 1 \[l-2N-{{H}_{2}}S{{O}_{4}}\]
D) 1 \[l-0.5N-{{H}_{2}}S{{O}_{4}}\]
Correct Answer: D
Solution :
\[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[1\text{ }\times \text{ }1\text{ }=\text{ }0.5\text{ }\times \text{ }2\text{ }\times \text{ }1\] as 0.5 N will give double the amount of H+ ions
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