Question

An aqueous solution of NaOH was made and its molar mass from the measurement of osmotic pressure at 27°C was found to be 25 g mol$^{-1}$.
Calculate the percentage dissociation of NaOH in this solution.

Hint:

In osmotic pressure calculations, remember that dissociation increases the effective number of particles. The van't Hoff factor \(i\) plays a key role in determining the extent of dissociation.

Answer 1

The osmotic pressure (\(\Pi\)) is given by the equation: \[ \Pi = i \cdot M \cdot R \cdot T \] Where: - \( i \) is the van't Hoff factor, - \( M \) is the molarity of the solution, - \( R \) is the gas constant, - \( T \) is the temperature in Kelvin. We know the molar mass of NaOH is 25 g/mol, and its molarity is calculated based on the mass of NaOH dissolved in the solution. From the osmotic pressure measurement, we can find \( i \), the degree of dissociation, and then calculate the percentage dissociation using the formula: \[ \text{Percentage dissociation} = \frac{(i - 1)}{i} \times 100 \]