A 0.13 g of a specimen containing MnO2 is treated with iodide ions. If iodine liberated requires 30.0 mL of 0.075M solution of Na2S2O3, the percentage of MnO2 in the mineral is:
A
75.3%
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B
85.3%
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C
95.3%
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D
none of the above
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Solution
The correct option is A75.3%
The following reactions takes place:
MnO2+I⊝→I2+Mn2+ I2+2Na2S2O3→Na2S4O6+2NaI
Milliequivalents of Na2SO3=(1×0.075)×30 Milliequivalents of Na2S2O3= milliequivalents of I2= milliequivalents of MnO2 ⇒WMnO2872×1000=(1×0.075)×30 ⇒ Percentage of MnO2=75.3%