Question

In the laboratory you dissolve 22.8 g of chromium(III) acetate in a volumetric flask and add water to a total volume of 250 mL. What is the molarity of the solution? M. What is the concentration of the chromium(III) cation? M. What is the concentration of the acetate anion?

Answer 1

Answer:

The molarity of the chromium(III) acetate solution 0.3982 M.

0.3982 M is the concentration of the chromium(III) cation.

1.1946 M is the concentration of the acetate anion.

Explanation:

$Molarity=\frac{\text{Mass of substance}}{\text{Molar mass of substance}\times \text{Volume of solution(L)}}$

Mass of chromium(III) acetate = 22.8 g

Molar mass of chromium(III) acetate = 229 g/mol

Total volume of the solution of chromium(III) acetate =  V = 250 mL = 0.250 L

1 mL = 0.001 L

The molarity of the chromium(III) acetate solution :

$[Cr(CH_3COO)_3]=\frac{22.8 g}{229 g/mol\times 0.250 L}=0.3982 M$

The molarity of the chromium(III) acetate solution 0.3982 M.

$Cr(CH_3COO)_3\rightarrow Cr^{3+}+3CH_3COO^-$

According to reaction, 1 mole of chromium(III) acetate gives 1 mole of chromium(III) ions and 3 moles of acetate ions.

So $[Cr^{3+}]=[Cr(CH_3COO)_3]=0.3982 M$

$[CH_3COO^-]=3\times [Cr(CH_3COO)_3]=3\times 0.3892 M=1.1946 M$

0.3982 M is the concentration of the chromium(III) cation.

1.1946 M is the concentration of the acetate anion.