How many grams of CH3OH must be added to water to prepare 420.0 mL of a solution that is 5.0 M CH3OH?
This all has to do with the definition of molarity. M = moles/L. It's that simple.
moles need for 420 mL solution of 5.0 M?
M x L = moles needed.
moles = grams/molar mass. Substitute moles needed and molar mass, solve for grams.
To determine the number of grams of CH3OH needed to prepare the solution, we need to use the molarity and volume of the solution.
First, let's understand the relationship between molarity, moles, and volume.
Molarity (M) is defined as moles of solute per liter of solution. In this case, the molarity is given as 5.0 M, meaning there are 5.0 moles of CH3OH per liter of solution.
We are given the volume of the solution as 420.0 mL. However, we need to convert it to liters in order to use the molarity properly. There are 1000 mL in 1 liter, so 420.0 mL is equal to 0.4200 liters.
Now, let's use the conversion factor between moles and liters to find the moles of CH3OH in the solution:
moles of CH3OH = molarity * volume (in liters)
= 5.0 M * 0.4200 L
= 2.1 moles
Next, we need to determine the molar mass of CH3OH, which consists of one carbon atom (C), four hydrogen atoms (H), and one oxygen atom (O). Using the atomic masses from the periodic table:
C: 12.01 g/mol
H: 1.01 g/mol (there are 4 hydrogen atoms)
O: 16.00 g/mol
Molar mass of CH3OH = (12.01 g/mol) + (1.01 g/mol * 4) + (16.00 g/mol)
= 32.04 g/mol
Finally, we can calculate the mass of CH3OH needed to prepare the solution:
mass of CH3OH = moles of CH3OH * molar mass of CH3OH
= 2.1 moles * 32.04 g/mol
= 67.25 grams
Therefore, you would need to add 67.25 grams of CH3OH to water to prepare 420.0 mL of a 5.0 M CH3OH solution.