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Molarity = moles of solute/Liters of solution

0.75 M KCl = moles KCl/2.25 Liters

= 1.6875 moles KCl (74.55 grams/1 mole KCl)

= 126 grams of KCl needed

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13y ago
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13y ago

74 grams

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11y ago

9.13

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Q: How many grams of KCl are required to prepare 2.25L of a 0.75m KCl solution?
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