A) \[10\,\,mL\]
B) \[20\,\,mL\]
C) \[5\,\,mL\]
D) \[15\,\,mL\]
Correct Answer: A
Solution :
Firstly molarity is converted into normality as Molarity\[\times \]mol. wt. \[=\] Normality \[\times \] Eq. wt. For\[{{H}_{2}}S{{O}_{4}}\] \[1\times 98=N\times 49\] \[N=\frac{98}{49}=2\] For\[NaOH\], \[1M=1N\] Now from normality equation \[{{H}_{2}}S{{O}_{4}}=NaOH\] \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[2\times {{V}_{1}}=1\times 20\] \[{{V}_{1}}=\frac{20}{2}=10\,\,mL\]of\[{{H}_{2}}S{{O}_{4}}\]
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