A 100 ml solution containing AgNO3 was treated with excess NaCl to completely precipitate the silver as AgCl. If 5.7 g AgCl was obtained, what was the concentration of Ag+ in the original solution?
A
0.03M
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B
0.05M
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C
0.12M
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D
0.30M
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E
0.40M
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Solution
The correct option is D0.40M The molecular weight of AgCl is 108+35.5=143.5 g/mol.
5.7 g AgCl corresponds to 5.7g143.5g/mol=0.03972 moles.
0.03972 moles of AgCl are obtained from 0.03972 moles of AgNO3. The volume of AgNO3 is 100 ml or 0.100 L.
[AgNO3]=0.03972mol0.100L=0.3972M≃0.40M
Hence, the concentration of Ag+ in the original solution was 0.40 M.