A) 8.0
B) 9.5
C) 12.5
D) 15.0
Correct Answer: C
Solution :
\[\underset{{{N}_{1}}{{V}_{1}}\,\,\,=}{\mathop{HCl\,\,\,\,\,\,}}\,\,\,\,\underset{{{N}_{2}}{{V}_{2}}}{\mathop{NaOH}}\,\] \[{{N}_{1}}\times 20ml=0.1\times 25\] \[{{N}_{1}}=\frac{0.1\times 25}{20}=0.125\]. If one litre HCl present in 0.125 Therefore in 10 ml \[\frac{0.125}{1000}\ \times \ 10\] = 12.5 The normality of conc. HCl is 12.5N.You need to login to perform this action.
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