The molar mass of the solute sodium hydroxide obtained from the measurement of osmotic pressure of its aqueous solution at 27^oC is 25g mol^–1. Therefore its ionization percentage in this solution is

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Detailed Solution

$NaOH \to {Na}^{+} + {OH}^{-} i = 1 + α i = \frac{Normal molar mass}{Observed molecular mass} = \frac{40}{25} = 1.6 1 + α = 1.6 \Rightarrow α = 0.6 The % dissociation = 0.6 \times 100 = 60 %$

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