A 1.50 g sample of an ore containing silver was dissolved, and all the ag+ ions were converted to 0.125 g ag2s. What was the percentage of silver in the ore?
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Answer:
The percentage of silver in the ore is 7.25%
Explanation:
Given data:
Weight of the sample ore = 1.50 g
Ag+ ions were converted to 0.125 g of Ag₂S.
To find: % of silver in the ore
Molecular weight of Ag₂S = 247.8 g/mol
Molecular weight of Silver = 107.9 g/mol
Weight of Ag present in 0.125 g of Ag₂S
= 0.125 * (fraction of silver in Ag₂S)
= 0.125 * [(107.9 * 2) / 247.8]
= 0.1088 g
∴ Percentage of silver in the ore
= (Weight of silver present in Ag₂S / Weight of the sample ore) * 100
= [0.1088 / 1.50] * 100
= 7.25
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