A) \[\text{40 mL}\]
B) \[\text{20 mL}\]
C) \[10\,mL\]
D) \[4\,mL\]
Correct Answer: A
Solution :
Equivalent weight of oxalic acid\[~=126/2\] (\[\because \]Basicity of oxalic acid\[=2\]) Normality of oxalic acid solution \[=\frac{6.3\times 1000}{63\times 250}\] \[=0.4\,N\] \[\underset{(Acid)}{\mathop{{{N}_{1}}{{V}_{1}}}}\,=\underset{(Base)}{\mathop{{{N}_{2}}{{V}_{2}}}}\,\] \[0.4\times 10\,mL=0.1\times {{V}_{2}}\] \[{{V}_{2}}=40\,mL\]
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