Asked by Anonymous
How many grams of solute are in 2.l L of 3.7Moles of Al(NO3)3
Answers
Answered by
DrBob222
I assume, although you don't strate it that way, that is 3.7M (as in molar and not mols).
Then mols Al(NO3)3 =- M x L = ?
and grams = mols x molar mass.
Then mols Al(NO3)3 =- M x L = ?
and grams = mols x molar mass.
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