Q. a 50g sample of Al(NO3)3 is dissolved in water to prepare 1500ml of solution. What is the concentration, in mol/L, of NO3 ions in the solution?
-The answer is 0.47 mol/L but I don't know how to solve it.
I am beginning to think you are answer mooching.
What is the concentration of the Al(NO3) in solution?
What is the concentration of the Nitrate ions?
I will be happy to critique your thinking.
To find the concentration of NO3 ions in the solution, we need to follow a series of steps:
Step 1: Calculate the number of moles of Al(NO3)3.
- First, determine the molar mass of Al(NO3)3.
- Molar mass of Al(NO3)3 = (1 x molar mass of Al) + (3 x molar mass of NO3)
- Molar mass of Al = 26.98 g/mol (atomic mass of aluminum)
- Molar mass of NO3 = (14.01 g/mol for nitrogen + 3 x 16.00 g/mol for oxygen)
- Calculate the molar mass of Al(NO3)3:
- Molar mass of Al(NO3)3 = (26.98 g/mol) + (3 x (14.01 g/mol + 3 x 16.00 g/mol))
- Multiply the molar mass by the mass of the sample to get the number of moles of Al(NO3)3:
- Moles of Al(NO3)3 = (mass of sample) / (molar mass of Al(NO3)3)
Step 2: Calculate the volume of the solution in liters.
- Convert the given volume from milliliters to liters:
- Volume of solution = 1500 ml = 1.5 L
Step 3: Calculate the concentration of NO3 ions.
- Divide the moles of Al(NO3)3 by the volume of the solution in liters:
- Concentration of NO3 ions (in mol/L) = (moles of Al(NO3)3) / (volume of solution in liters)
Putting it all together:
- Calculate the moles of Al(NO3)3:
- Moles of Al(NO3)3 = (50 g) / (molar mass of Al(NO3)3)
- Calculate the concentration of NO3 ions:
- Concentration of NO3 ions (in mol/L) = (moles of Al(NO3)3) / (volume of solution in liters)
Now, let's substitute the values given into the formula:
- Molar mass of Al(NO3)3 = (1 x 26.98 g/mol) + (3 x (14.01 g/mol + 3 x 16.00 g/mol)) = 212.99 g/mol
- Moles of Al(NO3)3 = (50 g) / (212.99 g/mol) ≈ 0.235 mol
- Volume of solution = 1.5 L
- Concentration of NO3 ions = (0.235 mol) / (1.5 L) ≈ 0.1567 mol/L
But remember, this is the concentration of Al(NO3)3, not just NO3 ions. Since there are three NO3 ions in one molecule of Al(NO3)3, we need to multiply the concentration by the factor of 3:
- Concentration of NO3 ions = 0.1567 mol/L x 3 = 0.4701 mol/L ≈ 0.47 mol/L
Thus, the concentration of NO3 ions in the solution is approximately 0.47 mol/L.