How many grams of barium acetate, Ba(CH3COO)2, must be dissolved to prepare 250. mL of a 0.218 M aqueous solution of the salt?
How many grams of chromium(III) sulfate, Cr2(SO4)3, must be dissolved to prepare 100. mL of a 0.123 M aqueous solution of the salt?
How many mL of a 0.167 M aqueous solution of calcium bromide, CaBr2, must be taken to obtain 8.51 grams of the salt? mL
The Correct Answer and Explanation is :
1. Grams of Barium Acetate (Ba(CH₃COO)₂) for 250 mL of 0.218 M Solution
Step 1: Calculate the moles of Ba(CH₃COO)₂ required.
The formula for molarity is:
[
Molarity (M) = \frac{moles\ of\ solute}{volume\ of\ solution\ in\ liters}
]
Given:
- Molarity ( M = 0.218 \, \text{M} )
- Volume ( V = 250 \, \text{mL} = 0.250 \, \text{L} )
Now, calculate the moles of Ba(CH₃COO)₂:
[
\text{moles} = M \times V = 0.218 \, \text{mol/L} \times 0.250 \, \text{L} = 0.0545 \, \text{mol}
]
Step 2: Calculate the mass of Ba(CH₃COO)₂.
The molar mass of Ba(CH₃COO)₂ can be calculated by adding the atomic masses of the elements:
- Ba: 137.33 g/mol
- C: 12.01 g/mol
- H: 1.008 g/mol
- O: 16.00 g/mol
Molar mass of Ba(CH₃COO)₂:
[
\text{Molar mass} = 137.33 + 2 \times (2 \times 12.01 + 3 \times 1.008 + 2 \times 16.00) = 137.33 + 2 \times 59.04 = 137.33 + 118.08 = 255.41 \, \text{g/mol}
]
Now, calculate the mass of Ba(CH₃COO)₂:
[
\text{mass} = \text{moles} \times \text{molar mass} = 0.0545 \, \text{mol} \times 255.41 \, \text{g/mol} = 13.9 \, \text{g}
]
Answer:
To prepare 250 mL of a 0.218 M solution of Ba(CH₃COO)₂, you need 13.9 grams of the salt.
2. Grams of Chromium(III) Sulfate (Cr₂(SO₄)₃) for 100 mL of 0.123 M Solution
Step 1: Calculate the moles of Cr₂(SO₄)₃ required.
Given:
- Molarity ( M = 0.123 \, \text{M} )
- Volume ( V = 100 \, \text{mL} = 0.100 \, \text{L} )
Calculate the moles of Cr₂(SO₄)₃:
[
\text{moles} = M \times V = 0.123 \, \text{mol/L} \times 0.100 \, \text{L} = 0.0123 \, \text{mol}
]
Step 2: Calculate the molar mass of Cr₂(SO₄)₃.
The molar mass of Cr₂(SO₄)₃ can be calculated by adding the atomic masses:
- Cr: 52.00 g/mol
- S: 32.07 g/mol
- O: 16.00 g/mol
Molar mass of Cr₂(SO₄)₃:
[
\text{Molar mass} = 2 \times 52.00 + 3 \times (32.07 + 4 \times 16.00) = 104.00 + 3 \times 96.07 = 104.00 + 288.21 = 392.21 \, \text{g/mol}
]
Now, calculate the mass of Cr₂(SO₄)₃:
[
\text{mass} = \text{moles} \times \text{molar mass} = 0.0123 \, \text{mol} \times 392.21 \, \text{g/mol} = 4.82 \, \text{g}
]
Answer:
To prepare 100 mL of a 0.123 M solution of Cr₂(SO₄)₃, you need 4.82 grams of the salt.
3. Volume of 0.167 M CaBr₂ Solution to Obtain 8.51 Grams
Step 1: Calculate the moles of CaBr₂ required.
Given:
- Molar mass of CaBr₂ = 40.08 g/mol (Ca) + 2 × 79.90 g/mol (Br) = 40.08 + 159.80 = 199.88 g/mol
- Mass of CaBr₂ = 8.51 g
Now, calculate the moles of CaBr₂:
[
\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{8.51 \, \text{g}}{199.88 \, \text{g/mol}} = 0.0426 \, \text{mol}
]
Step 2: Calculate the volume of 0.167 M CaBr₂ solution required.
Use the molarity equation:
[
M = \frac{\text{moles}}{\text{volume}}
]
Rearrange to solve for volume:
[
\text{volume} = \frac{\text{moles}}{M} = \frac{0.0426 \, \text{mol}}{0.167 \, \text{mol/L}} = 0.255 \, \text{L} = 255 \, \text{mL}
]
Answer:
To obtain 8.51 grams of CaBr₂, you need to take 255 mL of a 0.167 M aqueous solution.
Summary of Answers:
- 13.9 grams of Ba(CH₃COO)₂ for 250 mL of a 0.218 M solution.
- 4.82 grams of Cr₂(SO₄)₃ for 100 mL of a 0.123 M solution.
- 255 mL of a 0.167 M solution of CaBr₂ to obtain 8.51 grams of the salt.