Please help!!!! I need an answer quick!!!

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23-03-2020
Chemistry

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Please help!!!! I need an answer quick!!!

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joseaaronlara joseaaronlara · Answer 1 · 2020-03-24T00:19:22+01:00

Answer:

The answer to your question is 16.7 g of Cr(CH₃COO)₃

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

anfabba15 anfabba15 · Answer 2 · 2020-03-24T00:22:09+01:00

Answer:

16.7 g of acetate

Explanation:

Molarity (mol/L), a sort of concentration that indicates the moles of solute in 1 L of solution.

We have the solution's volume, so let's convert it to L

300 mL . 1L/1000mL = 0.3 L

Molarity = mol/ vol(L)

Molarity . vol(L) = moles → 0.243 M . 0.300L = 0.0729moles

Chromium (III) acetate = Cr(CH₃COO)₃ → Molar mass = 229 g/mol

We convert the moles to mass (moles . molar mass) = 0.0729 mol . 229 g/mol = 16.7 g