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you ned to make an aqeous solution of .139M aluminum nitrate for an experiment in a lab, using 300. ml volumetric flask. how much solid aluminum nitrate

User Timgavin
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1 Answer

20 votes
20 votes

Answer:

8.88 grams of aluminum nitrate should be weighted.

Step-by-step explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to use the definition of molarity to calculate the moles of aluminum nitrate as follows:


M=(n)/(V)\\\\n=M*V\\\\n=0.139mol/L*0.300L=0.0417mol

Now, since the molar mass of aluminum nitrate is 212.996 g/mol, we obtain the following mass:


m=0.0417mol*(212.996 g)/(1mol) \\\\m=8.88g

Therefore, 8.88 grams of aluminum nitrate should be weighted.

Regards!

User Ede
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