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How much potassinm hy.drexide is neede d to make 1,0O L of a 2 molar solution of ..... potassium hydvoxide?

How much potassinm hy.drexide is neede d to make 1,0O L of a 2 molar solution of ..... potassium hydvoxide?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Question:** How much potassium hydroxide is needed to make 1.00 L of a 2 molar solution of potassium hydroxide? **Explanation:** This handwritten note presents a problem in chemistry to calculate the amount of potassium hydroxide (KOH) required to prepare a 2 M (molar) solution with a total volume of 1.00 liter (L). To solve this problem, use the formula for molarity: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \] Given: - Molarity (M) = 2 mol/L - Volume (V) = 1.00 L First, calculate the number of moles of KOH required: \[ \text{Moles of KOH} = \text{Molarity} \times \text{Volume} \] \[ \text{Moles of KOH} = 2 \, \text{mol/L} \times 1.00 \, \text{L} = 2 \, \text{moles} \] Next, calculate the mass of KOH needed, using the molar mass of KOH (approximately 56.1 g/mol): \[ \text{Mass} = \text{moles} \times \text{molar mass} \] \[ \text{Mass} = 2 \, \text{moles} \times 56.1 \, \text{g/mol} = 112.2 \, \text{g} \] So, **112.2 grams of potassium hydroxide** is needed to make 1.00 L of a 2 molar solution of KOH.
Transcribed Image Text:**Question:** How much potassium hydroxide is needed to make 1.00 L of a 2 molar solution of potassium hydroxide? **Explanation:** This handwritten note presents a problem in chemistry to calculate the amount of potassium hydroxide (KOH) required to prepare a 2 M (molar) solution with a total volume of 1.00 liter (L). To solve this problem, use the formula for molarity: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \] Given: - Molarity (M) = 2 mol/L - Volume (V) = 1.00 L First, calculate the number of moles of KOH required: \[ \text{Moles of KOH} = \text{Molarity} \times \text{Volume} \] \[ \text{Moles of KOH} = 2 \, \text{mol/L} \times 1.00 \, \text{L} = 2 \, \text{moles} \] Next, calculate the mass of KOH needed, using the molar mass of KOH (approximately 56.1 g/mol): \[ \text{Mass} = \text{moles} \times \text{molar mass} \] \[ \text{Mass} = 2 \, \text{moles} \times 56.1 \, \text{g/mol} = 112.2 \, \text{g} \] So, **112.2 grams of potassium hydroxide** is needed to make 1.00 L of a 2 molar solution of KOH.
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