Question

The molar mass of the solute sodium hydroxide obtained from the measurement of osmotic pressure of its aqueous solution at ${27}^{\circ }C$ is 25 g/mol therefore, its ionization percentage in this solution is:

• A. 75
• B. 60
• C. 80
• D. 70

Answer 1

The correct option is B 60

$i=\frac{M_{ideal}}{M_o}=\frac{40}{25}=1.6\begin{array}{ccc}NaOH\to & N{a}^{+}+& O{H}^{-}\\ 1\text{mole}& 0& 0\\ 1-\alpha & \alpha & \alpha \end{array}$
Total = $1+\alpha$
Now, $i=1+\alpha$
or $\alpha =i-1$
or, $\alpha =1.6-1=0.6$
$\therefore$ Ionization percentage $=0.6\times 100=60\text{\%}$