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