A) \[1.6\times {{10}^{-3}}\]
B) 125
C) 2.5
D) None of these
Correct Answer: C
Solution :
[a] \[\overset{+4}{\mathop{SeO_{3}^{2-}}}\,+\overset{+5}{\mathop{BrO_{3}^{-}}}\,+{{H}^{+}}\to \overset{+6}{\mathop{SeO_{4}^{2-}}}\,+\overset{0}{\mathop{B{{r}_{2}}}}\,+{{H}_{2}}O\] [b] \[\overset{+5}{\mathop{BrO_{3}^{-}}}\,+\overset{+3}{\mathop{AsO_{2}^{-}}}\,+{{H}_{2}}O\to \overset{-1}{\mathop{B{{r}^{-}}}}\,+\overset{-5}{\mathop{AsO}}\,\,_{4}^{3-}+{{H}^{+}}\] In reaction [b] gm. Eq. of \[Br{{O}_{3}}^{-}=gm.\,eq.\]of \[As{{O}_{2}}^{-}\] \[{{n}_{BrO_{3}^{-}}}\times 6={{n}_{AsO_{2}^{-}}}\times 2=\frac{12.5}{1000}\times \frac{1}{25}\times 2={{10}^{-3}}\] \[{{n}_{BrO_{3}^{-}}}=\frac{{{10}^{-3}}}{6}\] In reaction [a] moles of \[BrO_{3}^{-}\]consumed \[=\frac{70}{1000}\times \frac{1}{60}-\frac{{{10}^{-3}}}{6}={{10}^{-3}}\] gm eq. of \[SeO_{3}^{2-}=gm.eq.\]of \[BrO_{3}^{-}\] \[{{n}_{SeO_{3}^{2-}}}\times 2={{10}^{-3}}\times 5;{{n}_{SeO_{3}^{2-}}}=2.5\times {{10}^{-3}}\]You need to login to perform this action.
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