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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

•Chemistry - DrBob222, Saturday, November 24, 2012 at 8:53pm
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.
4. Add more water until 600.0mL of solution is obtained
5. Mix

•Chemistry HELP - Anon, Sunday, November 25, 2012 at 2:01am
Isn't the total mass of the Al(NO3)3*9H2O around 247.18 g/mol?

  • Chemistry -

    Have I made an error?
    Al = about 27
    NO3 = about 62 and 3*62 = 186
    9H2O = 9*18 = about 162

    27 + 186 + 162 = about 375?

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