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Solid calcium iodide is slowly added to 175 ml of a ammonium fluoride solution until the concentration of calcium ion is 0.01
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Answer #1

Moles of calcium in solution = molarity×volume/1000

= 0.0122×175/1000

= 0.002135 moles

Moles of fluoride = 2×moles of calcium

= 2×0.002135 = 0.00427 moles

Concentration of fluoride = moles×1000/volume

= 0.00427×1000/175

= 0.0244 M

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