In basic medium, CrO2−4 oxidize S2O2−3 to form SO2−4 and itself changes to Cr(OH)−4. How many mL of 0.154 M CrO2−4 are required to react with 40 mL of 0.246 M S2O2−3 ?
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 S2O2−3=Milliequivalent of CrO2−4 40×0.246×8 = V×0.154×3 ∴V=0.154×340×0.246×8=170.4 mL Hence, (c) is the correct answer.