Answer:
0.95 L
Explanation:
Step 1: Given data
Concentration of the Mg(NO₃)₂ solution (C): 0.32 M (0.32 mol/L)
Mass of Mg(NO₃)₂ (solute): 45 g
Step 2: Calculate the moles corresponding to 45 g of Mg(NO₃)₂
The molar mass of Mg(NO₃)₂ is 148.33 g/mol.
45 g Mg(NO₃)₂ × 1 mol Mg(NO₃)₂ /148.33 g Mg(NO₃)₂ = 0.303 mol Mg(NO₃)₂
Step 3: Calculate the volume of solution that contains 0.303 moles of Mg(NO₃)₂
The concentration of the solution is 0.32 M, that is, there are 0.32 moles of Mg(NO₃)₂ per liter of solution.
0.303 mol Mg(NO₃)₂ × 1 L Solution / 0.32 mol Mg(NO₃)₂ = 0.95 L
