In order to make 500.0mL of a 1.835 M MgCl2 solution, then _______g of MgCl2 is needed.
Moles of MgCl2= (1.835M)(0.5L)=0.9175mol
Molecular Weight of MgCl2= 95.211 g/mol
(0.9175mol MgCl2)(95.211 g/mol)=
87.36g of MgCl2
Did I do this correct?
yes.
Yes, you have done the calculations correctly!
To determine the amount of MgCl2 needed to make a 500.0 mL solution of 1.835 M concentration, you need to find the number of moles of MgCl2 required. The formula to calculate moles is:
Moles = Concentration (M) × Volume (L)
So, in this case, the moles of MgCl2 are calculated as follows:
Moles of MgCl2 = 1.835 M × 0.5 L = 0.9175 mol
Next, you need to convert the moles of MgCl2 to grams using the molecular weight of MgCl2. The molecular weight of MgCl2 is 95.211 g/mol.
Mass of MgCl2 = Moles of MgCl2 × Molecular Weight of MgCl2
Mass of MgCl2 = 0.9175 mol × 95.211 g/mol = 87.36 g
So, in order to prepare 500.0 mL of a 1.835 M MgCl2 solution, you would need 87.36 grams of MgCl2.
Great job on your calculations!