How many grams of MGO are needed to prepare 2.5 L of a 0.5 molarity solution of MGO
It's MgO. Molarity is defined as mols solute/L solution.
If you want 2.5 L of 0.5 M solution, then you want how many mols? That's mols = M x L = ?.
Then how many grams do you need? That's mols = g/molar mass. You know molar mass (or can calculate it) and you know mols. Substitute and solve for grams MgO. Post your work if you get stuck.
To calculate the amount of MGO needed to prepare a solution, we first need to understand the relationship between molarity, volume, and the molecular weight of the compound. The formula used is:
Molarity (M) = moles of solute / liters of solution
In this case, we have the molarity (0.5 M) and the volume (2.5 L) of the solution and need to find the moles of MGO. To do this, we rearrange the formula:
moles of solute = Molarity * liters of solution
Now, we need to determine the molecular weight of MGO. Magnesium (Mg) has a molecular weight of 24.31 g/mol, and oxygen (O) has a molecular weight of 16.00 g/mol. Since there are two oxygen atoms present in MGO, we add the atomic masses together:
Molecular weight of MGO = 24.31 g/mol (Mg) + 2 * 16.00 g/mol (O)
Molecular weight of MGO = 24.31 g/mol + 32.00 g/mol
Molecular weight of MGO = 56.31 g/mol
Now, we can calculate the moles of MGO:
moles of MGO = Molarity * liters of solution
moles of MGO = 0.5 M * 2.5 L
moles of MGO = 1.25 mol
Finally, we can determine the mass of MGO needed:
mass of MGO = moles of MGO * molecular weight of MGO
mass of MGO = 1.25 mol * 56.31 g/mol
mass of MGO = 70.39 g
Therefore, you would need 70.39 grams of MGO to prepare 2.5 liters of a 0.5 molarity solution of MGO.