What is the molarity (M) of a 627 mL solution that contains 12.7 g of ethanol, CH3CH2OH?
Convert the grams to moles by using the molar mass then convert ml to liters
M= mol/liters
To calculate the molarity (M) of a solution, you need to know the amount of solute (in moles) and the volume of the solution (in liters).
To find the moles of ethanol in the solution, you need to convert the given mass of ethanol to moles. To do that, you will need the molar mass of ethanol (C2H5OH).
Now, let's break down the steps to determine the molarity (M) of the solution:
Step 1: Find the molar mass of ethanol (C2H5OH):
- The molar mass of C is 12.01 g/mol.
- The molar mass of H is 1.01 g/mol.
- The molar mass of O is 16.00 g/mol.
Therefore, the molar mass of ethanol (C2H5OH) is:
(2 * 12.01 g/mol) + (6 * 1.01 g/mol) + 16.00 g/mol = 46.07 g/mol.
Step 2: Convert the mass of ethanol (12.7 g) to moles:
- Divide the given mass of ethanol by its molar mass:
12.7 g / 46.07 g/mol ≈ 0.2758 mol.
Step 3: Convert the volume of the solution (627 mL) to liters:
- 1 L = 1000 mL, so 627 mL / 1000 mL/L = 0.627 L.
Step 4: Calculate the molarity:
- Use the formula: Molarity (M) = Moles (mol) / Volume (L).
M = 0.2758 mol / 0.627 L ≈ 0.4396 M.
Therefore, the molarity (M) of the solution containing 12.7 g of ethanol in a volume of 627 mL is approximately 0.4396 M.