Question

Calculate the osmotic pressure of $0.01M$ solution of $KCl$ having degree of disociation of $80\%$ at ${27}^{\circ }C$

• A. $0.03atm$
• B. $0.44atm$
• C. $0.12atm$
• D. $1.03atm$

Answer 1

The correct option is B $0.44atm$

For dissociation:
$KCl$ dissociates into two ions.
$\therefore n=2$
Degree of dissociation $\left(\alpha \right)$ :
$\alpha =\frac{i-1}{n-1}$
$0.8=\frac{i-1}{2-1}i=1.8$

Osmotic pressure is given by :

$\pi =iCRT$

$C\text{denotes concentration}=0.01MR\text{denotes ideal gas constant}=0.0821Latmmo{l}^{-1}{K}^{-1}T\text{denotes temperature}={27}^{\circ }C=300K$

$\pi =1.8\times 0.01\times 0.0821\times 300\pi =0.44atm$