Answer:
(b, c) \[{{H}_{2}}S{{O}_{4}}+2NaOH\to
N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O\]
\[\frac{{{M}_{1}}{{V}_{1}}}{{{n}_{1}}}({{H}_{2}}S{{O}_{4}})=\frac{{{M}_{2}}{{V}_{2}}}{{{n}_{2}}}(NaOH)\]
or \[\frac{0.1\times
{{V}_{1}}}{1}=\frac{0.1\times 1}{2}\]
or \[{{V}_{1}}=0.5L=500mL\]
Thus, number of moles of \[{{H}_{2}}S{{O}_{4}}\]
used = \[\frac{MV}{1000}\]
\[=\frac{0.1\times
500}{1000}=0.05\]
Number of moles of \[N{{a}_{2}}S{{O}_{4}}\]
formed will also be equal to 0.05.
Mass of \[N{{a}_{2}}S{{O}_{4}}=0.05\times
142=7.1g\]
\[\text{Molarity}\,\text{of}\,\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{=}\frac{\text{Number}\,\text{of}\,\text{moles}}{\text{Volume}\,\text{in}\,\text{litre}}\]
\[=\frac{0.05}{2}=0.025mol\,{{L}^{-1}}\]
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