how many grams of solid magnesium nitrate should be added to 5.0 grams of solid sodium nitrate in a 500 mL volumetric flask in order to produce an aqueous 0.25M solution of NO3- ions (once the flask is filled to the 500mL volume line)?
No idea where to start for this question.
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.
To solve this question, we need to use the concept of molarity and the equation:
Molarity (M) = moles of solute / volume of solution (in liters)
First, let's calculate the number of moles of sodium nitrate (NaNO3) in 5.0 grams:
1) Calculate the molar mass of sodium nitrate (NaNO3):
Na = 22.99 g/mol
N = 14.01 g/mol
O = 16.00 g/mol
Molar mass of NaNO3 = (22.99 g/mol) + (14.01 g/mol) + (3 x 16.00 g/mol) = 85.00 g/mol
2) Calculate the moles of sodium nitrate:
moles = mass / molar mass
moles of NaNO3 = 5.0 g / 85.00 g/mol
Next, since sodium nitrate (NaNO3) dissociates into two nitrate ions (NO3-), we need to multiply the moles of sodium nitrate by 2 to get the moles of nitrate ions:
moles of NO3- ions = (moles of NaNO3) x 2
Now, let's substitute the given molarity value (0.25 M) and calculate the volume of the solution (in liters):
0.25 M = moles of NO3- ions / volume of solution (in liters)
0.25 M = (moles of NaNO3 x 2) / 0.500 L
Rearranging this equation to solve for moles of NaNO3:
moles of NaNO3 = (0.25 M x 0.500 L) / 2
Now we can calculate the mass of solid magnesium nitrate (Mg(NO3)2) that needs to be added:
1) Calculate the molar mass of magnesium nitrate (Mg(NO3)2):
Mg = 24.31 g/mol
N = 14.01 g/mol
O = 16.00 g/mol
Molar mass of Mg(NO3)2 = (24.31 g/mol) + (2 x (14.01 g/mol)) + (6 x 16.00 g/mol) = 148.33 g/mol
2) Calculate the moles of magnesium nitrate:
moles of Mg(NO3)2 = (moles of NaNO3) x 148.33 g/mol
Finally, multiply the moles of magnesium nitrate by its molar mass to obtain the mass:
mass of Mg(NO3)2 = (moles of Mg(NO3)2) x 148.33 g/mol
This is the mass of solid magnesium nitrate that should be added to the volumetric flask to obtain an aqueous 0.25M solution of NO3- ions.
How many mols NO3^- do you want.
That's M x L = mols = 0.25 x 0.5L = 0.125
How many mols do you have in the NaNO3? That's mols = grams/molar mass = 5.0/85 = approx 0.06 but you need a more accurate answer that the estimate.
How many more mols do you need to be furnished by Mg(NO3)? That's approx 0.125 - 0.06 = approx 0.06 mols NO3^-
Since Mg(NO3)2 furnishes 2 NO3^- for each 1 mol Mg(NO3)2 you need only 0.03 mols Mg(NO3)2. How many grams is that? That's grams = mols Mg(NO3)2 x molar mass Mg(NO3)2.