A) 5.0
B) 0.5
C) 2.5
D) 10.0
Correct Answer: C
Solution :
Equivalents of NaOH \[\text{=}\frac{\text{Weight}\,\text{of}\,\text{NaOH}}{\text{Eq}\,\text{Weight}\,\text{of}\,\text{NaOH }\!\!\times\!\!\text{ volume}\,\text{of}\,\text{solution}}\] \[=\frac{20}{40\times 1}=0.5\,N\] Let the volume of \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]used is V. \[\therefore \] Equivalents of \[{{H}_{2}}S{{O}_{4}}={{N}_{1}}V\] [For\[{{H}_{2}}S{{O}_{4}};N{{}_{1}}=2M=2\times 0.1=0.2\,N\]] Equivalents of acid = Equivalent of base \[0.2\,NV=0.5\,N\] \[V=\frac{0.5}{0.2}=2.5\]
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