Asked Dec 1, 2019

You are asked to prepare a 0.8500 M solution of aluminum nitrate. You find that you have only 50.00 g of the solid. What is the maximum volume of solution that you can prepare?

Volume =?


Expert Answer

Step 1

Molarity: Dividing moles of solute by the volume of solution in Liter.


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Molesof solute Molarity Volume of solution in L

Step 2

First find the mole of Aluminum nitrate that present in 50.0 g of Aluminum nitrate solid.

Aluminum nitrate molar mass = 212.996 g/mol


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Mass Mole = Molar mass 50.0g -=0.23mol 212.996g

Step 3

Volume of solution that will makes 0.23 mol Al...


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Moles of solute Molarity= Volume of solution in L Molesof solute Volume of solution in L=! Molarity 0.23mol 0.850 M 0.276L -276 mL


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