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Standard XII

Chemistry

Types of Redox Reactions

A mixture of ...

Question

A mixture of 0.02 mole of $K B r O_{3}$ and 0.01 mole KBr was treated with excess of KI and acidified. Determine the volume of 0.1 M $N a_{2} S_{2} O_{3}$ solution required to consume the liberated iodine.

120 mL

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12 mL

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240 mL

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60 mL

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Solution

The correct option is A 120 mL
$B r O_{3}^{-} + 5 B r^{-} + 6 H^{+} \to 3 B r_{2} + 3 H_{2} O$
$I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a_{2} S_{4} O_{6} + 2 N a I$
$B r_{2} \to I_{2} \to N a_{2} S_{2} O_{3}$
Amount of $N a_{2} S_{2} O_{3}$ reacting $= 0.006 * 2 = 0.012$ mole
$0.1 \times V \times 10^{- 3} = 0.012$
Here, V is the volume of $N a_{2} S_{2} O_{3}$ in ml.
$V = \frac{0.012 m l}{0.1 \times 10^{- 3}} = \frac{0.012 m l}{10^{- 4}} = 120 m l$

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Similar questions

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A mixture of 0.02 mole of KBrO3 and 0.01 mole KBr was treated with excess of KI and acidified. Determine the volume of 0.1 M Na2S2O3 solution required to consume the liberated iodine.

The correct option is A 120 mLBrO−3+5Br−+6H+→3Br2+3H2OI2+2Na2S2O3→2Na2S4O6+2NaIBr2→I2→Na2S2O3Amount of Na2S2O3 reacting =0.006∗2=0.012 mole0.1×V×10−3=0.012 Here, V is the volume of Na2S2O3 in ml.V=0.012ml0.1×10−3=0.012ml10−4=120 ml

A mixture of 0.02 mole of $K B r O_{3}$ and 0.01 mole KBr was treated with excess of KI and acidified. Determine the volume of 0.1 M $N a_{2} S_{2} O_{3}$ solution required to consume the liberated iodine.