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3 April, 00:18

How many grams of kcl are necessary to prepare 1.50 liters of a 0.500-m solution of kcl?

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  1. 3 April, 00:30
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    1.50L of 0.500m KCl

    1.50 (0.500/1 L) =.75 mol KCl

    .75 mol KCl (74.55 g/1 mol) = 55.1925 g KCl
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