A) 10 g
B) 80 g
C) 40 g
D) 20 g
E) None of these
Correct Answer: E
Solution :
\[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[{{N}_{1}}\](oxalic acid)\[=2\times 0.5N\] \[1\times 50={{N}_{2}}\times 25\Rightarrow {{N}_{2}}=2N\] Normality\[=\frac{wt.}{eq.wt.}\times \frac{1000}{V(mL)}\] \[\Rightarrow \]\[2=\frac{wt.}{40}\times \frac{1000}{50}\] weight = 4g
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