How many grams of methanol, CH3OH, are there in 150.0 mL of a 0.30 M aqueous solution of methanol
M = mols/L
Substitute and solve for mols.
Then mols = grams/molar mass. Substitute and solve for grams. About 1.5g (but an estimate)?
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
To find the number of grams of methanol, CH3OH, in 150.0 mL of a 0.30 M aqueous solution, you'll need to follow these steps:
Step 1: Determine the number of moles of methanol in the solution.
- Molarity (M) is defined as moles of solute per liter of solution.
- In this case, the molarity (M) is given as 0.30 M, so there are 0.30 moles of CH3OH in every liter of solution.
- First, convert the volume from milliliters (mL) to liters (L). There are 1000 mL in 1 L, so 150.0 mL is equal to 0.1500 L.
- Multiply the molarity by the volume (in liters) to find the number of moles of CH3OH.
Number of moles = Molarity x Volume
Number of moles = 0.30 mol/L x 0.1500 L
Step 2: Calculate the molar mass of methanol.
- Methanol, CH3OH, has a molar mass equal to the sum of the atomic masses of each element:
C: 12.01 g/mol, H: 1.01 g/mol, O: 16.00 g/mol, and H: 1.01 g/mol
- Molar mass of CH3OH = (12.01 g/mol x 1) + (1.01 g/mol x 4) + (16.00 g/mol + 1.01)
- Molar mass of CH3OH = 32.04 g/mol
Step 3: Calculate the number of grams of CH3OH.
- Multiply the number of moles of CH3OH (from Step 1) by the molar mass of CH3OH (from Step 2) to get the number of grams.
Number of grams = Number of moles x Molar mass
Number of grams = (0.30 mol/L x 0.1500 L) x 32.04 g/mol
By substituting the values into the equation, you can solve for the number of grams of CH3OH.