A) \[12c{{m}^{3}}\]
B) \[10c{{m}^{3}}\]
C) \[25c{{m}^{3}}\]
D) \[10.5c{{m}^{3}}\]
Correct Answer: B
Solution :
When \[0.1\text{ }N\text{ }NaOH\]is used, \[\underset{(For\,HCl)}{\mathop{{{N}_{1}}{{V}_{1}}}}\,=\underset{(For\,NaOH)}{\mathop{{{N}_{2}}{{V}_{2}}}}\,\] \[0.2N\times {{V}_{1}}=50\times 0.1N\] \[{{V}_{1}}=\frac{50\times 0.1}{0.2}=25c{{m}^{2}}\] When \[0.5N\text{ }KOH\]is used, \[{{N}_{1}}{{V}_{1}}\] = \[{{N}_{3}}{{V}_{3}}\] (For remaining \[HCl\]) (For \[KOH\]) \[0.2N\times 25=0.5N\times {{V}_{3}}\] \[{{V}_{3}}=\frac{0.2\times 25}{0.5}\] \[=10c{{m}^{3}}\]
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