You need to make an aqueous solution of 0.139M aluminum nitrate, Al(NO₃)₃ for an experiment in the lab, using a 300. mL volumetric flask. How much solid aluminum nitrate (molar mass = 213.012 g) should you add to the flask?

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Quizplus · Accepted Answer

To find the mass of aluminum nitrate needed, use the formula: mass = molarity (M) × volume (L) × molar mass (g/mol). Convert 300 mL to L (0.300 L), then multiply by the molarity (0.139 M) and the molar mass of aluminum nitrate (213.012 g/mol): 0.139 M × 0.300 L × 213.012 g/mol = 8.88 g.

You Need to Make an Aqueous Solution of 0