Multiple Choice
Determine the molarity of a solution prepared by dissolving 11.5 g of NaOH to form a 250.0 mL solution. (Molar mass of NaOH = 40.00 g/mol)
6. Chemical Quantities & Aqueous Reactions
Molarity
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How many grams of CH₃OH must be added to water to prepare 150 mL of a solution that is 2.00 M CH₃OH?
A
9.60 grams
B
3.00 grams
C
12.00 grams
D
6.00 grams
Verified step by step guidance1
Start by understanding the concept of molarity (M), which is defined as the number of moles of solute per liter of solution. In this problem, the molarity of CH₃OH is given as 2.00 M.
Calculate the number of moles of CH₃OH required for the solution. Use the formula: \( \text{Moles of CH}_3\text{OH} = \text{Molarity} \times \text{Volume of solution in liters} \). Convert the volume from milliliters to liters by dividing by 1000.
Determine the molar mass of CH₃OH. The molecular formula CH₃OH consists of carbon (C), hydrogen (H), and oxygen (O). Calculate the molar mass by adding the atomic masses: \( \text{Molar mass of CH}_3\text{OH} = 12.01 \text{ g/mol (C)} + 3 \times 1.01 \text{ g/mol (H)} + 16.00 \text{ g/mol (O)} \).
Use the number of moles calculated in step 2 and the molar mass from step 3 to find the mass of CH₃OH needed. Apply the formula: \( \text{Mass of CH}_3\text{OH} = \text{Moles of CH}_3\text{OH} \times \text{Molar mass of CH}_3\text{OH} \).
Verify your calculations to ensure accuracy. Check each step for correct unit conversions and arithmetic operations to confirm the mass of CH₃OH required for the solution.
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