A) 0.2
B) 0.1
C) 0.01
D) 10
Correct Answer: A
Solution :
\[\text{CaC}{{\text{O}}_{\text{3}}}\] is decomposed as : \[\underset{Mol\,wt\,=\,100\,}{\mathop{\text{CaC}{{\text{O}}_{3}}}}\,\xrightarrow{{}}CaO+\underset{Mol.\,wt\,=\,44(X)}{\mathop{C{{O}_{2}}}}\,\] So, the (X) will be \[\text{C}{{\text{O}}_{\text{2}}}\text{.}\] Now (X) is passed into an aqueous solution containing one mole of \[\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\] as \[CO{{}_{2}}+\underset{\text{Sodium}\,\text{carbonate}\,\text{(1}\,\text{mole)}}{\mathop{N{{a}_{2}}C{{O}_{3}}}}\,+{{H}_{2}}O\xrightarrow{{}}\] \[\underset{\text{Sodium}\,\text{bicarbonat}\,\text{(2}\,\text{moles)}}{\mathop{2NaHC{{O}_{3}}}}\,\] \[\because \] No. of moles of \[\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\] when 100g \[\text{CaC}{{\text{O}}_{3}}\] is decomposed = 2 \[\therefore \] No. of moles of \[\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\] when 10 g \[\text{CaC}{{\text{O}}_{\text{3}}}\] is decomposed \[\text{=}\,\frac{2\times 10}{100}=0.2\,\text{mole}\]
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