How do you prepare 600.0mL of 0.48M Al(NO3)3*9H2O?
Is this correct?
1. Measure out 71g of solution
2. Put approx. 300.0mL of water into container
3. Add the 71g of solution to the water and mix
4. Add more water until 600.0mL of solution is obtained
5. Mix
I would make some corrections, one major and two or three minor.
1. Measure out 71g of solution
71 g is not right
2. Put approx. 300.0mL of water into container
3. Add the xxg of solution(solute to the water and mix dissolve the solute completely.
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4. Add more water until 600.0mL of solution is obtained
5. Mix
Isn't the total mass of the Al(NO3)3*9H2O around 247.18 g/mol?
Al = about 27
NO3 = about 62*3 = 186
9H2O = 9*18 = 162
total is about 375 or did I make an error?
To prepare 600.0mL of 0.48M Al(NO3)3*9H2O solution, you need to follow the steps below:
1. Start by calculating the required amount of Al(NO3)3*9H2O using the given concentration and volume.
- Molarity (M) = moles of solute / liters of solution
- Number of moles = Molarity × volume (in liters)
- Given: Molarity = 0.48M and Volume = 0.600L
- Moles = 0.48M × 0.600L = 0.288 moles
2. Determine the molar mass of Al(NO3)3*9H2O.
- Aluminum (Al): atomic mass = 26.98 g/mol
- Nitrogen (N): atomic mass = 14.01 g/mol
- Oxygen (O): atomic mass = 16.00 g/mol
- Hydrogen (H): atomic mass = 1.01 g/mol
- Multiply the atomic masses by the respective number of atoms present in the formula (Al(NO3)3*9H2O).
- Aluminum (1) = 26.98 g/mol
- Nitrogen (1) = 14.01 g/mol
- Oxygen (12) = 192.00 g/mol
- Hydrogen (18) = 18.18 g/mol
- Total molar mass = 26.98 + 14.01 + 192.00 + 18.18 = 251.17 g/mol
3. Calculate the required mass of Al(NO3)3*9H2O.
- Mass = Number of moles × Molar mass
- Mass = 0.288 moles × 251.17 g/mol = 72.36 grams
4. Measure out 72.36 grams of the Al(NO3)3*9H2O compound.
5. Pour approximately 300.0mL of water into a container.
6. Add the 72.36 grams of the Al(NO3)3*9H2O compound to the water and mix well to dissolve it. It may be necessary to heat the solution slightly to help the solid dissolve faster.
7. Once the solid is completely dissolved, add more water to the container until the total volume reaches 600.0mL.
8. Mix the solution thoroughly to ensure uniformity.
Your revised steps should be as follows:
1. Measure out 72.36 grams of the Al(NO3)3*9H2O compound.
2. Pour approximately 300.0mL of water into a container.
3. Add the 72.36 grams of the Al(NO3)3*9H2O compound to the water and mix well.
4. Add more water to the container until the total volume reaches 600.0mL.
5. Mix the solution thoroughly to ensure uniformity.
Following these steps should allow you to successfully prepare 600.0mL of 0.48M Al(NO3)3*9H2O solution.