Question

$2\times {10}^{-3}M$ solution of $NaCl$ having degree of disociation of $90\%$ at ${27}^{\circ }C$ has osmotic pressure equal to :

• A. $0.5atm$
• B. $9.4atm$
• C. $0.094atm$
• D. $0.94atm$

Answer 1

The correct option is C $0.094atm$

For dissociation:
$NaCl$ dissociates into two ions.
$\therefore n=2$
Degree of dissociation $\left(\alpha \right)$ :
$\alpha =\frac{i-1}{n-1}$
$0.9=\frac{i-1}{2-1}i=1.9$

Osmotic pressure is given by :

$\pi =iCRT$

$C\text{denotes concentration}=2\times {10}^{-3}MR\text{denotes ideal gas constant}=0.0821Latmmo{l}^{-1}{K}^{-1}T\text{denotes temperature}={27}^{\circ }C=300K$

$\pi =1.9\times 2\times {10}^{-3}\times 0.0821\times 300\pi =0.094atm$