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