A) \[30.6\,%\]
B) \[40.6\,%\]
C) \[20.6\,%\]
D) \[50\,%\]
Correct Answer: C
Solution :
\[Meq.\,\,of\,\,{{H}_{2}}S{{O}_{4}}+Meq.of\,\,S{{O}_{3}}=Meq.of\,\,NaOH\] \[\therefore \,\,\,\,\,\,\,\frac{\left( 0.5-x \right)}{98/2}\times 1000+\frac{x}{80/2}\times 1000=26.7\times \,0.4\]\[\therefore ~~x\,\,=\,\,0.103\] \[\therefore ~~%\, of \,S{{O}_{3}} = \frac{0.103}{0.5} \times 100 = 20.6\,%\]You need to login to perform this action.
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