# Chemistry doubt

$$0.5\text{ g}$$ 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?

Note by Shivam Mishra
1 year, 10 months ago

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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 :

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

$M=\frac{5}{1000}$

As molecular weight of sulphite ion is $$80$$, percentage by mass is :

$\frac{0.005 × 80}{0.5} × 100$ $=80 %$

- 1 year, 9 months ago

Is the information complete?

- 1 year, 10 months ago

Yes

- 1 year, 10 months ago

80%

- 1 year, 8 months ago

Thanks @Utkarsh Dwivedi

- 1 year, 9 months ago

Tell me is this question based on molarity

- 1 year, 10 months ago

Yes,but it can also be solved using milliequivalents.

- 1 year, 10 months ago