A) 1.9 g, \[0.28\,mol\,{{L}^{-1}}\]
B) 13.6 g, \[0.28\,mol\,{{L}^{-1}}\]
C) 1.9 g, \[0.14\,mol\,{{L}^{-1}}\]
D) 13.6 g, \[0.14\,mol\,{{L}^{-1}}\]
Correct Answer: A
Solution :
\[\underset{-}{\mathop{\underset{100\,m\,mol}{\mathop{Ca{{(OH)}_{2}}}}\,}}\,+\underset{-}{\mathop{\underset{14\,m\,mol}{\mathop{N{{a}_{2}}S{{O}_{2}}}}\,}}\,\xrightarrow[{}]{{}}\underset{14\,m\,mol}{\mathop{\underset{{}}{\mathop{CaS{{O}_{4}}}}\,}}\,+\underset{28\,m\,mol}{\mathop{\underset{{}}{\mathop{2NaOH}}\,}}\,\] |
\[{{W}_{CaS{{O}_{4}}}}=14\times {{10}^{-3}}\times 136=1.9gm\] |
\[[O{{H}^{-}}]=\frac{28}{100}=0.28M.\] |
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