Answer:
The answer to your question is 16.7 g of Cr(CH₃COO)₃
Step-by-step explanation:
Data
Concentration = 0.243 M
Volume = 300 ml
mass of Chromium (III) acetate = ?
Process
1.- Write the formula of the compound
Cr(CH₃COO)₃
2.- Calculate the molar mass of the compound
Cr(CH₃COO)₃ = 52 + (6 x 12) + (9 x 1) + (6 x 16)
= 52 + 72 + 9 + 96
= 229 g
3.- Determine the number of moles
Molarity = moles /volume
Moles = Molarity x volume
Moles = 0.243 x 0.300
Moles = 0.0729
4.- Calculate the grams of chromium (III) acetate
229 g ------------------ 1 mol
x ----------------- 0.0729 mol
x = (0.0729 x 229) /1
x = 16.7 g of Cr(CH₃COO)₃