\(0.5\text{ g}\) sample of a sulphite salt was dissolved in \(200\text{ ml}\) and \(20\text{ ml}\) of this solution required \(10\text{ ml}\) of \(0.02\text{ M}\) acidified permanganate solution.Find the percentage by mass of sulphite in sulphite ore?

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## Comments

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TopNewestMy 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 :

\[M ×2×\frac{20}{200} = \frac{10}{1000} × 0.02 × 5\]

\[M=\frac{5}{1000}\]

As molecular weight of sulphite ion is \(80\), percentage by mass is :

\[\frac{0.005 × 80}{0.5} × 100\]

\[=80 %\]Log in to reply

Is the information complete?

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Yes

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@Deeparaj Bhat @Prakhar Bindal please help!!!!

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Tell me is this question based on molarity

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Yes,but it can also be solved using milliequivalents.

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Thanks @Utkarsh Dwivedi

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80%

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