Question

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

Answer 1

<b>I would make some corrections, one major and two or three minor.</b>

1. Measure out 71g of solution
<b>71 g is not right</b>

2. Put approx. 300.0mL of water into container
3. Add the <b>xx</b>g of solution<b>(solute</b> to the water and mix <b>dissolve the solute completely.</b>
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4. Add more water until 600.0mL of solution is obtained
5. Mix

Answer 2

Isn't the total mass of the Al(NO3)3*9H2O around 247.18 g/mol?

Answer 3

Al = about 27
NO3 = about 62*3 = 186
9H2O = 9*18 = 162
total is about 375 or did I make an error?