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A 500 mL potassium chloride solution was prepared by dissolving

potassium chloride in distilled water. 25.0 mL of the solution was
titrated with 0.300 M silver nitrate solution. 28.90 mL of silver
nitrate solution was required to reach the end point in the titration.
What is the number of moles of potassium chloride present in the
500.0 mL solution?

𝐾𝐢𝑙(π‘Žπ‘ž) + 𝐴𝑔𝑁𝑂3(π‘Žπ‘ž) β†’ 𝐴𝑔𝐢𝑙(𝑠) + 𝐾𝑁𝑂3(π‘Žπ‘ž)

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1 answer
  1. moles silver nitrate=.3*.0288
    moles KCL= moles silver nitrate=.0864
    now number moleKCl in 500ml= .0964*500/25=...

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    bobpursley

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