0.5 g sample of a sulphite salt was dissolved in 200 ml and 20 ml of this solution required 10 ml of 0.02 M acidified permanganate solution.Find the percentage by mass of sulphite in sulphite ore?
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My solution :
Let the number of moles of sulphite ions be M. Clearly on oxidation the sulphite ion changes to sulphate so the oxidation number changes from 4+ to 6+, hence the valence factor is 2. The product of moles and valence factor gives the equivalents. The valence factor permagnate solution in acidified solution is 5. Equate the equivalents as :
M×2×20020=100010×0.02×5
M=10005
As molecular weight of sulphite ion is 80, percentage by mass is :
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
My solution :
Let the number of moles of sulphite ions be M. Clearly on oxidation the sulphite ion changes to sulphate so the oxidation number changes from 4+ to 6+, hence the valence factor is 2. The product of moles and valence factor gives the equivalents. The valence factor permagnate solution in acidified solution is 5. Equate the equivalents as :
M×2×20020=100010×0.02×5
M=10005
As molecular weight of sulphite ion is 80, percentage by mass is :
0.50.005×80×100 =80%
Is the information complete?
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Yes
@Deeparaj Bhat @Prakhar Bindal please help!!!!
Tell me is this question based on molarity
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Yes,but it can also be solved using milliequivalents.
Thanks @Utkarsh Dwivedi
80%