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How many grams of KCl are needed to prepare 0.5 M solution in 500 mL of water? (
How many grams of KCl are needed to prepare 0.5 M solution in 500 mL of water? (2023)
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Practice Questions
1 question
Q1
How many grams of KCl are needed to prepare 0.5 M solution in 500 mL of water? (2023)
14.9 g
7.45 g
29.8 g
3.73 g
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Molarity = moles/volume(L). Moles = 0.5 mol/L * 0.5 L = 0.25 mol. Mass = moles * molar mass = 0.25 mol * 74.5 g/mol = 18.625 g.
Questions & Step-by-step Solutions
1 item
Q
Q: How many grams of KCl are needed to prepare 0.5 M solution in 500 mL of water? (2023)
Solution:
Molarity = moles/volume(L). Moles = 0.5 mol/L * 0.5 L = 0.25 mol. Mass = moles * molar mass = 0.25 mol * 74.5 g/mol = 18.625 g.
Steps: 5
Show Steps
Step 1: Understand that molarity (M) is the number of moles of solute per liter of solution.
Step 2: Identify the desired molarity, which is 0.5 M, and the volume of the solution, which is 500 mL (or 0.5 L).
Step 3: Calculate the number of moles needed using the formula: Moles = Molarity * Volume. So, Moles = 0.5 mol/L * 0.5 L = 0.25 mol.
Step 4: Find the molar mass of KCl. The molar mass of KCl is approximately 74.5 g/mol.
Step 5: Calculate the mass of KCl needed using the formula: Mass = Moles * Molar Mass. So, Mass = 0.25 mol * 74.5 g/mol = 18.625 g.
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