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0.800 M KOH (1mol/2.5L)(56.1g KOH/1mol)

It's set up stoichiometrically, but hard to show that here... 0.800 M KOH / 2.5 L x 56.1g


Your answer should come out 17.952 g KOH. If you're following sig figs, then your answer should come out 18. g KOH
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14y ago
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10y ago

The answer is 14,0264 g KOH.

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Q: Calculate the number of grams of solute that would be needed to make 2.5 L of 0.800 M KOH?
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