Question

A student weighs out a 4.31 g sample of Al(CH3COO)3, transfers it to a 100 mL volumetric flask, adds enough water to dissolve it and then adds water to the 100 mL tick mark. What is the Molarity of aluminum acetate in the resulting solution?

Answer 1

$C= \frac{m}{M \cdot V}$

$C$ - molarity of the solution, mol/l; $m$ - mass of the solute, g; $M$ - molar mass of the solute, g/mol; $V$ - volume of the solution, l.

$M(Al(CH_3COO)_3) \approx 1 \cdot 26.98 + 3 \cdot(2 \cdot 12.01 + 3 \cdot 1.01 + 2 \cdot 16.00) = 204.13 \; g/mol$

$C= \frac{4.31}{204.13 \cdot 0.1} \approx 0.21 \; mol/l$

Answer: 0.21 mol/l.