a solution is made by diluting 175 mL of 0.238 M aluminum nitrate, Al(NO3)3, solution with water to a final vlume of 5.00x 10^2 mL.calculate the following.
a. the molarity of aluminum nitrate, aluminum ion, and nitrate ion in the diluted solution
B. the number of moles of nitrate ion in the original solution.
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3.5
To calculate the molarity of a solution, we use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
a. Molarity of aluminum nitrate (Al(NO3)3) in the diluted solution:
Step 1: Convert the volume of the solution from mL to L.
Volume of solution = 5.00 x 10^2 mL = 500 mL = 500/1000 = 0.5 L
Step 2: Calculate the moles of aluminum nitrate:
Moles of aluminum nitrate = Molarity x Volume (in liters)
Moles of aluminum nitrate = 0.238 M x 0.5 L = 0.119 mol
Step 3: Calculate the molarity of aluminum nitrate in the diluted solution:
Molarity of aluminum nitrate = Moles of aluminum nitrate / Volume of solution (in liters)
Molarity of aluminum nitrate = 0.119 mol / 0.5 L = 0.238 M
b. To find the number of moles of nitrate ion in the original solution:
Step 1: Calculate the number of moles of aluminum nitrate in the original solution:
Moles of aluminum nitrate = Molarity x Volume (in liters)
Moles of aluminum nitrate = 0.238 M x 0.175 L = 0.04165 mol
Step 2: Since the formula for aluminum nitrate is Al(NO3)3, there are three nitrate ions (NO3-) for every one aluminum nitrate molecule.
Therefore, the number of moles of nitrate ion in the original solution is:
Moles of nitrate ion = 3 x Moles of aluminum nitrate = 3 x 0.04165 mol = 0.12495 mol (approximately 0.125 mol)
So, the number of moles of nitrate ion in the original solution is approximately 0.125 mol.
Sure! To calculate the molarity of aluminum nitrate, aluminum ion, and nitrate ion in the diluted solution, we can use the equation:
M1V1 = M2V2
Where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
a. Molarity of aluminum nitrate in the diluted solution:
M1 = 0.238 M (given)
V1 = 175 mL = 0.175 L (since 1 L = 1000 mL)
V2 = 500 mL = 0.5 L
Plugging in these values into the equation, we have:
0.238 M * 0.175 L = M2 * 0.5 L
Solving for M2, we get:
M2 = (0.238 M * 0.175 L) / 0.5 L
= 0.0833 M
Therefore, the molarity of aluminum nitrate in the diluted solution is 0.0833 M.
To calculate the molarity of aluminum ion and nitrate ion in the diluted solution, we need to know the stoichiometry of Al(NO3)3.
The balanced equation for the dissociation of aluminum nitrate is:
Al(NO3)3 → Al3+ + 3NO3-
This means that for each mole of aluminum nitrate, we get one mole of aluminum ion and three moles of nitrate ion.
Given that the molarity of aluminum nitrate is 0.0833 M, the molarity of aluminum ion (Al3+) in the diluted solution will also be 0.0833 M.
Since there are three nitrate ions (NO3-) for every one aluminum ion (Al3+), the molarity of nitrate ion in the diluted solution will be three times the molarity of aluminum nitrate:
Nitrate ion molarity = 3 * 0.0833 M = 0.2499 M (approximately 0.250 M)
b. To calculate the number of moles of nitrate ion in the original solution, we need to use the equation:
moles = molarity * volume
Molarity of nitrate ion = 0.238 M (given)
Volume of nitrate ion solution = 175 mL = 0.175 L (since 1 L = 1000 mL)
Plugging the values into the equation, we have:
moles = 0.238 M * 0.175 L
= 0.04165 moles
Therefore, there are approximately 0.04165 moles of nitrate ion in the original solution.