A) 176.66 ml
B) 156.6 ml
C) 116.0 ml
D) 196.1 ml
Correct Answer: A
Solution :
[a] Normality of 3% \[N{{a}_{2}}C{{\text{O}}_{3}}\]. \[N=\frac{3\times 100}{53\times 100}=0.566N\] For \[{{H}_{2}}S{{O}_{4}}\,sol.{{N}_{1}}=0.1,{{V}_{1}}=100\,mL\] For \[N{{a}_{2}}C{{O}_{3}}\,sol.\,{{N}_{2}}=0.566.{{N}_{2}}?\] Now apply \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[{{V}_{2}}=\frac{{{N}_{1}}{{V}_{1}}}{{{N}_{2}}}=\frac{0.1\times 100mL}{0.566}=176.66mL\]
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