Question

Osmotic pressure of a solution is 12 atm. What is the concentration of NaCl solution which is isotonic to the given solution at 900 K? Given \( R = 0.082 \, \text{L-atm K}^{-1} \text{mol}^{-1} \) (Assume 100% dissociation)

• A. 0.4878 M
• B. 0.0243 M
• C. 0.243 M
• D. 0.04878 M

Hint:

When calculating osmotic pressure, the van't Hoff factor \( i \) accounts for the number of particles produced from the dissociation of solute. For NaCl, \( i = 2 \) due to complete dissociation into Na\(^+\) and Cl\(^-\).

Answer 1

The Correct Option is C

Step 1: Understand the definition of osmotic pressure.
Osmotic pressure \( \Pi \) is given by the formula: \[ \Pi = i M R T \] where: - \( i \) is the van't Hoff factor (which is 2 for NaCl due to 100% dissociation), - \( M \) is the molarity of the solution, - \( R \) is the gas constant, - \( T \) is the temperature in Kelvin. For isotonic solutions, the osmotic pressure of NaCl solution should be equal to the osmotic pressure of the given solution. Hence, we can write: \[ \Pi_{\text{NaCl}} = \Pi_{\text{given}} \] \[ i M R T = 12 \, \text{atm} \]
Step 2: Substitute known values.
For NaCl solution, \( i = 2 \), \( R = 0.082 \, \text{L-atm K}^{-1} \text{mol}^{-1} \), and \( T = 900 \, \text{K} \), so the equation becomes: \[ 2 M (0.082) (900) = 12 \] Solving for \( M \): \[ M = \frac{12}{2 \times 0.082 \times 900} = \frac{12}{147.6} = 0.243 \, \text{M} \] Thus, the concentration of the NaCl solution is \( 0.243 \, \text{M} \).