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. How many grams of potassium chloride, KCl, are needed to prepare 0.630 L of a 2.50 M solution of potassium chloride? (4.0 points) (Points :4)
1.575 M
117 g
2.54 m
23.19 g

  • chemistry - ,

    solve by
    molarity =no. of moles divide by volume in litres.......

    so let the mass be in x grms...
    then :
    2.50*0.630=x grms
    x grms of KCl =1.575 grams is the answer

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