A) \[600\,\,mL\]
B) \[300\,\,mL\]
C) \[500\,\,mL\]
D) \[200\,\,mL\]
Correct Answer: C
Solution :
Key Idea: Use the following formula to find volume of\[NaOH\]. \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] Given, \[{{N}_{1}}=\]normality of\[NaOH=0.4\,\,M\] \[{{V}_{1}}=\]volume of\[NaOH=?\] \[{{N}_{3}}=\]normality of\[NaOH=?\] \[{{V}_{2}}=\]volume of\[{{H}_{2}}S{{O}_{4}}=200\,\,mL\] \[\therefore \] \[0.4\times {{V}_{1}}=(2\times 0.5)\times 200\] \[(\because \]it is diprotic acid) or \[{{V}_{1}}=\frac{2\times 0.5\times 200}{0.4}\] \[=500\,\,mL\]
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