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Question

A metal ore weighing 6.125 g was roasted in air to get 5.725 g of the oxide. The oxide was then put into sulfuric acid and dried. This gave us a white solid of mass 7.725 g. The final solid is a

A
Sulphide
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B
Sulphate
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C
Sulphite
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D
None of the above
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Solution

The correct option is B Sulphate
Number of moles of sulphur present in the ore = 6.1255.7253216=140
Mass of metal = 6.125140×32=5.325g
Mass of new ion in the final white solid = 7.7255.3251/40=96g/mol
Let's assume the formula of the ion to be SOx
Hence, 96=32+16×x64=16×xx=4
Hence, the ion is SO24.

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