A) 11.10 mL
B) 16.65 Ml
C) 22.20 mL
D) 5.55 mL
Correct Answer: D
Solution :
| Normality \[\text{=}\frac{\text{wt }\!\!%\!\!\text{ }\,\text{ }\!\!\times\!\!\text{ }\,\text{density }\,\text{ }\!\!\times\!\!\text{ 10}}{\text{eq}\text{.wt}\text{.}}\] |
| \[=\frac{98\times 1.8\times 10}{49}=36\,N\] |
| \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] |
| \[36\times V=0.2\times 1000\] |
| \[V=\frac{0.2\times 1000}{36}=5.55\,\,mL.\] |
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