How many mols of Al(NO3)^2 do you need? That's mols - M x L = ?
Now convert mols Al(NO3)2 to grams. That's g = mols x molar mass = ?
What mass of aluminium nitrate (Al(NO3)3) would be required to prepare 2.000 L of a 0.0160 M aqueous solution of this salt?
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