Question

If the observed and normal osmotic pressures of $1\%$ NaCl solution are 5.7 and 3.0 atm, the degree of dissociation of NaCl is:

• A. 0.9
• B. 1
• C. 0.57
• D. 0.3

Answer 1

The correct option is A 0.9

$\begin{array}{c}i=\frac{\text{observed colligative property}}{\text{normal colligative property}}=\frac{5.7}{3}=1.9\\ \\ \text{Degre of dissociation}\alpha =\frac{i-1}{n-1}=\frac{0.9}{2-1}=0.9\\ \\ \text{Hence, option}A\text{is correct.}\end{array}$