Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
![**Question:**
Which aqueous solution would have the lowest osmotic pressure at 298K?
a) 0.04 M CaCl₂
b) 0.05 M NaBr
c) 0.1 M LiBr
d) 0.2 M K₂SO₄
**Explanation:**
To determine which solution has the lowest osmotic pressure, we can use the formula for osmotic pressure:
\[ \Pi = iMRT \]
Where:
- \(\Pi\) = osmotic pressure,
- \(i\) = van't Hoff factor (number of particles the compound dissociates into),
- \(M\) = molarity,
- \(R\) = ideal gas constant,
- \(T\) = temperature in Kelvin.
Calculate the effective concentration (\(i \times M\)) for each solution:
- **For CaCl₂**: \(i = 3\) (CaCl₂ dissociates into 3 ions: Ca²⁺ and 2 Cl⁻)
Effective concentration = \(3 \times 0.04 = 0.12\)
- **For NaBr**: \(i = 2\) (NaBr dissociates into 2 ions: Na⁺ and Br⁻)
Effective concentration = \(2 \times 0.05 = 0.10\)
- **For LiBr**: \(i = 2\) (LiBr dissociates into 2 ions: Li⁺ and Br⁻)
Effective concentration = \(2 \times 0.1 = 0.20\)
- **For K₂SO₄**: \(i = 3\) (K₂SO₄ dissociates into 3 ions: 2 K⁺ and SO₄²⁻)
Effective concentration = \(3 \times 0.2 = 0.60\)
The solution with the lowest effective concentration of ions will have the lowest osmotic pressure. Thus, the answer is **b) 0.05 M NaBr**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff753bfa8-1ffa-4ebf-8e17-943d9e514202%2Fe9ef6da4-f1cf-44ea-9cf4-886379045a17%2Fkpbiy9n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
Which aqueous solution would have the lowest osmotic pressure at 298K?
a) 0.04 M CaCl₂
b) 0.05 M NaBr
c) 0.1 M LiBr
d) 0.2 M K₂SO₄
**Explanation:**
To determine which solution has the lowest osmotic pressure, we can use the formula for osmotic pressure:
\[ \Pi = iMRT \]
Where:
- \(\Pi\) = osmotic pressure,
- \(i\) = van't Hoff factor (number of particles the compound dissociates into),
- \(M\) = molarity,
- \(R\) = ideal gas constant,
- \(T\) = temperature in Kelvin.
Calculate the effective concentration (\(i \times M\)) for each solution:
- **For CaCl₂**: \(i = 3\) (CaCl₂ dissociates into 3 ions: Ca²⁺ and 2 Cl⁻)
Effective concentration = \(3 \times 0.04 = 0.12\)
- **For NaBr**: \(i = 2\) (NaBr dissociates into 2 ions: Na⁺ and Br⁻)
Effective concentration = \(2 \times 0.05 = 0.10\)
- **For LiBr**: \(i = 2\) (LiBr dissociates into 2 ions: Li⁺ and Br⁻)
Effective concentration = \(2 \times 0.1 = 0.20\)
- **For K₂SO₄**: \(i = 3\) (K₂SO₄ dissociates into 3 ions: 2 K⁺ and SO₄²⁻)
Effective concentration = \(3 \times 0.2 = 0.60\)
The solution with the lowest effective concentration of ions will have the lowest osmotic pressure. Thus, the answer is **b) 0.05 M NaBr**.
Expert Solution
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Step 1: Osmotic pressure:
Osmotic pressure is defined as the pressure that would have to be applied to a solution to prevent the flow of solvent into the solution across a semipermeable membrane. This pressure is directly proportional to the concentration of solute particles in the solution. The osmotic pressure can be calculated using the following formula:
П = iMRT,
where:
i = Van't Hoff factor
П= osmotic pressure (atm)
M = molar concentration of solute particles (mol/L)
R = gas constant (0.0821 L.atm/mol.K)
T = temperature (K)
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