How many grams of ethanol, C2H5OH, are required to make a 7.1M solution using 1.50 L of water
You want how many mols. That is M x L = 7.1 x 1.5 = ?
mols = grams/molar mass. You know molar mass ethanol and mols, solve for grams.
To calculate the number of grams of ethanol required to make a 7.1M solution using 1.50 L of water, we need to understand the concept of molarity and use the equation:
Molarity (M) = moles of solute / volume of solution in liters
First, let's find the number of moles of ethanol that would be required. We know the desired molarity (7.1M) and the volume of the solution in liters (1.50 L).
Rearranging the equation above, we have:
moles of solute = Molarity (M) x volume of solution (L)
moles of solute = 7.1 mol/L x 1.50 L = 10.65 moles
Now, we need to convert moles of ethanol to grams. To do this, we need the molar mass of ethanol (C2H5OH). The molar mass of each element can be found from the periodic table:
Carbon (C): 12.01 g/mol
Hydrogen (H): 1.01 g/mol (there are two H atoms in ethanol)
Oxygen (O): 16.00 g/mol
Total molar mass of ethanol (C2H5OH): (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + 16.00 g/mol
Total molar mass of ethanol (C2H5OH) = 46.07 g/mol
Now, we can calculate the mass of ethanol needed:
Mass = moles x molar mass
Mass = 10.65 moles x 46.07 g/mol
Mass = 490.90 grams
Therefore, approximately 490.90 grams of ethanol (C2H5OH) are required to make a 7.1M solution using 1.50 L of water.