To solve this question, we need to use the concept of molarity (M). Molarity is defined as the number of moles of solute per liter of solution. The formula for molarity is:
Molarity (M) = (moles of solute) / (volume of solution in liters)
First, we need to calculate the number of moles of sodium nitrate (NaNO3) in the given mass. To do this, we need the molar mass of NaNO3, which can be calculated by adding the atomic masses of sodium (Na), nitrogen (N), and three oxygen (O) atoms:
Molar mass of NaNO3 = (1 * atomic mass of Na) + (1 * atomic mass of N) + (3 * atomic mass of O)
Next, we can calculate the moles of NaNO3 by dividing the given mass by the molar mass.
moles of NaNO3 = (mass of NaNO3) / (molar mass of NaNO3)
Now, we can calculate the volume (in liters) of the solution. The given volume is 500 mL, which is equivalent to 0.5 L.
Next, we need to determine the moles of NO3- ions required for a 0.25M solution. The molar ratio between NaNO3 and NO3- ions is 1:1, so the moles of NO3- ions should be the same as the moles of NaNO3.
Finally, we can calculate the mass of magnesium nitrate required to produce the desired molarity of NO3- ions. The molar mass of magnesium nitrate (Mg(NO3)2) can be calculated by adding the atomic masses of magnesium (Mg), two nitrogen (N), and six oxygen (O) atoms.
Once we have the molar mass of Mg(NO3)2, we can use the molar mass and the calculated moles of NO3- ions to find the mass of magnesium nitrate.
Let's calculate the values step by step:
Step 1: Calculate the molar mass of NaNO3.
Atomic mass of Na = 22.99 g/mol
Atomic mass of N = 14.01 g/mol
Atomic mass of O = 16.00 g/mol
Molar mass of NaNO3 = (1 * 22.99) + (1 * 14.01) + (3 * 16.00) = 85.00 g/mol
Step 2: Calculate the moles of NaNO3.
moles of NaNO3 = 5.0 g / 85.00 g/mol
Step 3: Calculate the volume of the solution.
Given volume = 500 mL = 0.5 L
Step 4: Determine the moles of NO3- ions required.
moles of NO3- ions = moles of NaNO3
Step 5: Calculate the molar mass of Mg(NO3)2.
Atomic mass of Mg = 24.31 g/mol
Atomic mass of N = 14.01 g/mol
Atomic mass of O = 16.00 g/mol
Molar mass of Mg(NO3)2 = (1 * 24.31) + (2 * 14.01) + (6 * 16.00) = 148.31 g/mol
Step 6: Calculate the mass of Mg(NO3)2.
mass of Mg(NO3)2 = (moles of NO3- ions) * (molar mass of Mg(NO3)2)
Now, you have all the information to solve the problem. Plug in the calculated values into the appropriate equations and solve for the mass of solid magnesium nitrate that should be added.