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A mixture of 0.02 mole of KBrO(3) and 0....

A mixture of 0.02 mole of `KBrO_(3)` and `0.001` mole of `KBr` was treated with excess of KI and acidified. The volume of 0.01M `Na_(2)S_(2)O_(3)` solution required to consume the liberated iodine will be :

A

1000 mL

B

1200 mL

C

1500 mL

D

800 mL

Text Solution

Verified by Experts

The correct Answer is:
B
`BrO_(3)^(-)+6I^(-)rarr3I_(2)+Br^(-)`
moles of `I_(2)=3xx`moles of `KBrO_(3)`
`:. ` moles of `I_(2)=0.02xx3=0.06`
Eq of `I_(2)`=Eq of Hypo
`0.06xx2=0.1xxV`
V=1.2 L = 1200 mL.
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