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 defined by the equation:\[\Pi = i \cdot M \cdot R \cdot T\]In this equation:- \( i \) represents the van't Hoff factor.- \( M \) denotes the molarity of the solution.- \( R \) signifies the gas constant.- \( T \) indicates the temperature in Kelvin.Given the molar mass of NaOH as 25 g/mol, its molarity is determined by the mass of NaOH dissolved. Osmotic pressure measurements allow for the determination of \( i \), which is the degree of dissociation. The percentage dissociation can then be calculated using the formula:\[\text{Percentage dissociation} = \frac{(i - 1)}{i} \times 100\]