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Question

In basic medium, CrO24 oxidize S2O23 to form SO24 and itself changes to Cr(OH)4. How many mL of 0.154 M CrO24 are required to react with 40 mL of 0.246 M S2O23 ?

A
200 mL
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B
156.4 mL
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C
170.4 mL
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D
190.4 mL
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Solution

The correct option is C 170.4 mL
Milliequivalent of S2O23=Milliequivalent of CrO24
40×0.246×8 = V×0.154×3
V=40×0.246×80.154×3=170.4 mL
Hence, (c) is the correct answer.

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