Question

If the observed and normal osmotic pressures of a KCl solution are 5.85 and 3.20 atm, the degree of dissociation of KCl is:

• A. 1
• B. 0.082
• C. 0.82
• D. 0.28

Answer 1

Answer: (C)

0.82

$\begin{array}{cc}i& =\frac{\text{observed colligative propety}}{\text{normal colligative property}}\\ & =\frac{5.85}{3.2}\\ & =1.828\end{array}$

$\begin{array}{c}\text{Degree of dissociation}\alpha =\frac{i-1}{n-1}\\ \begin{array}{cc}\alpha & =\frac{0.82}{2-1}\\ & =0.82\end{array}\\ \\ \text{Hence option}C\text{is correct.}\end{array}$