Final answer:
To make a 2.00 L solution of 0.150 M magnesium nitrate, you need 44.4981 g of Mg(NO₃)₂, calculated using the molarity formula and the molar mass of magnesium nitrate.
Step-by-step explanation:
To calculate the mass of magnesium nitrate, Mg(NO₃)₂, needed to prepare a 2.00 L solution with a concentration of 0.150 M you use the formula:
Mass (g) = Molar concentration (M) × Volume (L) × Molar mass (g/mol)
First, calculate the number of moles needed:
Moles = 0.150 M × 2.00 L = 0.300 moles
Then, find the molar mass of Mg(NO₃)₂:
Molar mass of Mg(NO₃)₂ = (24.305 g/mol for Mg) + (2 × (14.007 g/mol for N + 3 × 15.999 g/mol for O))
Molar mass of Mg(NO₃)₂ = 24.305 + 2 × (14.007 + 3 × 15.999) = 148.327 g/mol
Finally, calculate the mass:
Mass of Mg(NO₃)₂ = 0.300 moles × 148.327 g/mol = 44.4981 g
So, you would need 44.4981 g of magnesium nitrate to prepare the solution.