Question
Asked Oct 16, 2019
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Calculate the volume in milliliters of a 0.72/molL aluminum sulfate solution that contains 200.g of aluminum sulfate (Al2(SO4)3)

Be sure your answer has the correct number of significant digits.

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Expert Answer

Step 1

It is required to calculate the volume of 0.72 mol/L aluminium sulfate solution containing 200 g of aluminium sulfate. This can be done as,

Mass
No. of moles
Molar Mass
Mass 200 g
Molar mass of Al2(SO4)3 = 2 x 27 + 3 x [32 16 x 4]
Molar mass of Al2(SO4)3 = 54 + 288 342 g/mol
200
No. of moles
342
No. of moles
0.584 mols
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Mass No. of moles Molar Mass Mass 200 g Molar mass of Al2(SO4)3 = 2 x 27 + 3 x [32 16 x 4] Molar mass of Al2(SO4)3 = 54 + 288 342 g/mol 200 No. of moles 342 No. of moles 0.584 mols

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Step 2

The volume can be ca...

Moles
Molarity
Volume in litres
0.584 mol
0.72 mol/L =
Volume in litres
Volume 0.812 L = 812 mL
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Moles Molarity Volume in litres 0.584 mol 0.72 mol/L = Volume in litres Volume 0.812 L = 812 mL

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