Place ~0.8 g Fe(NO3)3·9 H2O into a 10 mL volumetric flask and dissolve to the volumetric line with 0.1 M HNO3(aq). (Make sure you pre-rinse the volumetric flask with the 0.1 M HNO3(aq) solution before making the solution. Also, 9 waters of hydration (·9H2O) are present in this chemical, and must be included in the formula weight calculation.) Record the exact mass of iron nitrate used and show the calculation for the solution concentration in your observations. Before lab, create a plan for the dilution of the Fe(NO3)3(aq) solution created above with DI water. This plan should be clearly outlined in your ELN. Create enough 0.0020 M Fe(NO3)3

How do I go about solving for the dilution plan?

You can take 1mL of your stock solution in a 100mL volumetric flask and complete to volume.

Explanation:

You need to create a 0.00200M solution of Fe(NO₃)₃. First, you have to obtain the concentration of the first solution you made. That is:

0.8g Fe(NO₃)₃.9H₂O × (1mol / 403.9972g) =

0.0020 moles of Fe(NO₃)₃.9H₂O = Moles of Fe(NO₃)₃

In 10mL = 0.010L:

0.0020 moles of Fe(NO₃)₃ / 0.010L = 0.20M Fe(NO₃)₃

This is the concentration of your stock solution, as you want to obtain a 0.0020M solution, you dilution factor must be:

0.20M / 0.0020M = 100

That means you need to dilute your stock solution 100 times.

You can make this dilution, for example,

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"