A) \[2.5\]
B) \[5.0\]
C) \[10.0\]
D) \[20.0\]
Correct Answer: B
Solution :
Molecular weight of \[{{H}_{2}}S{{O}_{4}}=98\] Equivalent weight of \[{{H}_{2}}S{{O}_{4}}=\frac{98}{2}=49\] \[Molarity\times mol.\text{ }wt.=normality\times eq.\text{ }wt.\] \[1\times 98={{N}_{1}}\times 49\] \[{{N}_{1}}=\frac{98}{49}=2\] Similarly, for \[NaOH\] \[1N=1M\] \[{{N}_{2}}=1\] Normality equation, \[\underset{Acid}{\mathop{{{N}_{1}}{{V}_{1}}}}\,=\underset{Base}{\mathop{{{N}_{2}}{{V}_{2}}}}\,\] \[2\times {{V}_{1}}=1\times 10\] \[{{V}_{1}}=5mL\] So, \[5mL\] \[1M\] \[{{H}_{2}}S{{O}_{4}}\] is required to neutralize \[10mL\]of \[1\text{ }M\]\[NaOH\] solution.
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