M = mols/L
Substitute and solve for mols.
Then mols = grams/molar mass. Substitute and solve for grams. About 1.5g (but an estimate)?
How many grams of methanol, CH3OH, are there in 150.0 mL of a 0.30 M aqueous solution of methanol
2 answers
Recall that molarity is just moles of solute per liter of solution:
M = n / V
Since we're given the volume and molarity, we solve for the n:
0.30 M = n / 0.15 L
n = 0.3 * 0.15
n = 0.045 mol CH3OH
The problem asks for the mass of CH3OH. To get the mass, we multiply the number of moles by the molar mass of CH3OH.
To get the molar mass, get a periodic table and add the individual masses of the elements in the chemical formula:
CH3OH : 1*12 + 1*3 + 1*16 + 1*1 = 32 g/mol
Thus,
0.045 mol * 32 g/mol = 1.44 g CH3OH
Hope this helps :3
M = n / V
Since we're given the volume and molarity, we solve for the n:
0.30 M = n / 0.15 L
n = 0.3 * 0.15
n = 0.045 mol CH3OH
The problem asks for the mass of CH3OH. To get the mass, we multiply the number of moles by the molar mass of CH3OH.
To get the molar mass, get a periodic table and add the individual masses of the elements in the chemical formula:
CH3OH : 1*12 + 1*3 + 1*16 + 1*1 = 32 g/mol
Thus,
0.045 mol * 32 g/mol = 1.44 g CH3OH
Hope this helps :3