What volume of solvent should be added to
36.8 g of CH3OH to produce a 0.500 M solution of CH3OH?
To calculate the volume of solvent needed, you first need to determine the number of moles of CH3OH in 36.8 g using its molar mass.
The molar mass of CH3OH is approximately 32.04 g/mol:
C: 1 carbon atom x 12.01 g/mol = 12.01 g/mol
H: 4 hydrogen atoms x 1.01 g/mol = 4.04 g/mol
O: 1 oxygen atom x 16.00 g/mol = 16.00 g/mol
Total molar mass: 12.01 g/mol + 4.04 g/mol + 16.00 g/mol = 32.04 g/mol
Next, use the equation:
moles = mass / molar mass
moles of CH3OH = 36.8 g / 32.04 g/mol = 1.149 mol
Now, you need to find the volume of solvent to achieve a 0.500 M solution. "M" represents molarity, which is defined as moles of solute per liter of solution.
Molarity = moles of solute / volume of solution
Rearranging the equation gives:
Volume of solution = moles of solute / Molarity
Volume of solution = 1.149 mol / 0.500 mol/L = 2.298 L
Therefore, you need to add approximately 2.298 liters of solvent to 36.8 g of CH3OH to produce a 0.500 M solution of CH3OH.
To find the volume of solvent needed to prepare a certain concentration of a solution, you can use the formula:
Volume of solvent = (mass of solute) / (molarity of solution × molar mass of solute)
Given:
Mass of solute (CH3OH) = 36.8 g
Molarity of solution = 0.500 M (mol/L)
Molar mass of CH3OH = 32.04 g/mol
Plugging in the values into the formula:
Volume of solvent = (36.8 g) / (0.500 mol/L × 32.04 g/mol)
First, calculate the molar mass of CH3OH: 0.5 mol/L × 32.04 g/mol = 16.02 g
Then, divide the mass of solute by the molar mass of CH3OH: 36.8 g / 16.02 g = 2.299 L
Therefore, you would need to add approximately 2.299 liters of solvent to the 36.8 g of CH3OH to produce a 0.500 M solution of CH3OH.
How many moles do you have of CH3OH? That is mols = grams/molar mass = ?
Then M = moles/L soln.
You know M and you know mols, solve for L.
That L will be the total volume; I suppose the author of the problem expects you to subtract the total volume - volume CH3OH in the 36.8 g but there is no density given and the volumes are not additive. Frankly, I don't think that part of the problem can be solved.