A) 0.107 M
B) 0.126 M
C) 0.214 M
D) - 0.428 M
Correct Answer: A
Solution :
\[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[{{N}_{NaOH}}={{M}_{NaOH}}=0.164\] \[\Rightarrow 25\times N=32.63\times 0.164\] \[N=\frac{32.63\times .164}{25}=0.214\text{N}\] But \[{{N}_{{{H}_{2}}S{{O}_{4}}}}=2\times {{M}_{{{H}_{2}}S{{O}_{4}}}}\] \[\Rightarrow \] \[M=\frac{\text{Normality}}{2}=\frac{0.214}{2}=0.107\]
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