A) \[12\,\,c{{m}^{3}}\]
B) \[10\,\,c{{m}^{3}}\]
C) \[21.0\,\,c{{m}^{3}}\]
D) \[16.2\,\,c{{m}^{3}}\]
Correct Answer: B
Solution :
No. of equivalent of \[HCl\] remaining after adding \[50\,c{{m}^{3}}\]of \[0.1\,N\,NaOH=\frac{0.2\times 50-0.1\times 50}{100}\] \[=\frac{0.5}{100}\] \[\therefore \] Volume of \[0.5\text{ }N\text{ }KOH\] required \[=\frac{0.5}{100}eq\] \[\equiv \frac{V\times 0.5}{1000}\] \[V=\frac{0.5}{100}\times \frac{1000}{0.5}\] \[=10\,\,c{{m}^{3}}\]
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