The molar mass of the solute sodium hydrdoxide obtained from the measurement of the osmotic pressure of its aqueous solution at `27^(@)C` is `25 g mol^(-1)`. Therefore its ionization percentage in this solution is

Options

A. 75
B. 60
C. 80
D. 70

Answer

B

Explanation

`i=(M_(c))/(M_(0))=(40)/(25)=1.6`
`{:(NaOHrarrNa^(+)+OH^(-)),("1 mol"),(1-alpha" "alpha" "alpha" Total "=1+alpha):}`
`i=1+alpha or alpha=i-1=0.6=60%`