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Ca(OH)2 = 74.093 g/mol

25g / 74.093 g/mol = 0.3374 mol Ca(OH)2

0.3374 mol / 0.5 mol/L = 0.6748 L Ca(OH)2

which is equal to approximately 675 mL, or 680 rounded up.

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Q: What will the volume of 0.50 M solution be if it contains 25 grams of calicum hydroxide?
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