A) 1 N
B) 0.1 N
C) 10 N
D) 11 N
Correct Answer: B
Solution :
Molarity of \[{{H}_{2}}S{{O}_{4}}=0.5\] Normality of \[{{H}_{2}}S{{O}_{4}}\,\,({{N}_{1}})=0.5\times 2=1\] \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[1\times 1={{N}_{2}}\times 10\] or \[{{N}_{2}}=\frac{1}{10}=0.1N.\]You need to login to perform this action.
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