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How many grams of KCl are needed to prepare 300 mL of a 0.25 M KCl solution?

Question:

How many grams of KCl are needed to prepare 300 mL of a 0.25 M KCl solution?

Molarity:

The compound needed to form the solution is KCl (potassium chloride) and the concentration of the solution needed to prepare is 0.25 M (expressed in molar concentration). Knowing the molar concentration of the solution as a unit of concentration in chemistry indicates the number of moles of KCl needed per liter of the solution.

Answer and Explanation: 1

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We need to calculate the number of moles of KCl needed to prepare the solution.

{eq}\rm Moles=\left(Molarity\right)\times \left(Volume\right) {/eq}

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Calculating Molarity and Molality Concentration

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Chapter 8 / Lesson 4
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What is molarity? What is molality? Compare molarity vs molality by viewing molality and molarity formulas, how to calculate molarity and molality, and examples.


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