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20 " mL of " (M)/(60) KBrO3 was reacted ...

20 " mL of " `(M)/(60)` `KBrO_3` was reacted with a sample of `SeO_(3)^(2-)` 20 The `Br_(2)` thus evolved was removed and the excess of `NaAsO_(2)` The reaction involved are
`SeO_(3)^(2-)+BrO_(3)^(ɵ)+H^(o+)toSeO_(4)^(2-)+Br_(2)+H_(2)O` .. (i)
`BrO_(3)^(ɵ)+ASO_(2)^(ɵ)+H_(2)OtoBr^(ɵ)+AsO_(4)^(3-)+H^(o+)` ..(ii)
`[Mw(SeO_(3)^(2-))=79+48=127gmol^(-1)]`
Q. n-factors of `BrO_(3)^(ɵ)` ion in equations (i) and (ii) respectively are

A

10,6

B

5,6

C

6,10

D

6,5

Text Solution

Verified by Experts

The correct Answer is:
B
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20 " mL of " (M)/(60) KBrO_3 was reacted with a sample of SeO_(3)^(2-) 20 The Br_(2) thus evolved was removed and the excess of NaAsO_(2) The reaction involved are SeO_(3)^(2-)+BrO_(3)^(ɵ)+H^(o+)toSeO_(4)^(2-)+Br_(2)+H_(2)O .. (i) BrO_(3)^(ɵ)+ASO_(2)^(ɵ)+H_(2)OtoBr^(ɵ)+AsO_(4)^(3-)+H^(o+) ..(ii) [Mw(SeO_(3)^(2-))=79+48=127gmol^(-1)] Q. Which of the following is true (T) or false (F)?

20 " mL of " (M)/(60) KBrO_3 was added to a sample of SeO_3^(2-) , The bromine evolved was removed and the excess of KBrO_3 was titrated with 5.1 " mL of " (M)/(25) solution of NaAsO_2 . Calculate the amound of SeO_3^(2-) and balance the equation. SeO_3^(2-)+BrO_3^(ɵ)+H^(o+)toSeO_4^(2-)Br_2+H_2O BrO_3^(ɵ)+AsO_2^(ɵ)+H_2OtoBr^(ɵ)+AsO_4^(3-)+H^(o+) (Br=80, K=39, As=75, Se=79)

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