A) 50 mL
B) 75 mL
C) 12.5 mL
D) 25 mL
Correct Answer: D
Solution :
\[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] For \[N{{a}_{2}}C{{O}_{3}}({{N}_{2}})=2\times 0.1N\] \[{{N}_{1}}\times 25=2\times 0.1\times 30\] \[{{N}_{1}}=\frac{2\times 0.1\times 30}{25}=0.24\] For titration with NaOH \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[0.24\times {{V}_{1}}=0.2\times 30\] \[{{V}_{1}}=\frac{0.2\times 30}{0.24}=25mL\]
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