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how many grams of solid KCl are needed to prepare 250ml of 0.235 M solution?

2007-10-27 14:52:25 · 1 answers · asked by Foster W 1 in Science & Mathematics ➔ Chemistry

1 answers

Let the solution be called S.

Atomic weights: K=39 Cl=35.5 KCl=74.5

250mLS x 0.235molKCl/1000mLS x 74.5gKCl/1molKCl = 4.38g KCl to three sig figs.

2007-10-27 15:03:04 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋

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