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Question

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
0.03 M
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B
0.05 M
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C
0.12 M
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D
0.30 M
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E
0.40 M
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Solution

The correct option is D 0.40 M
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=0.3972M0.40M

Hence, the concentration of Ag+ in the original solution was 0.40 M.

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