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 MK₂SO4

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**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**.
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
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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