Question

A 0.5 M solution of weak electrolyte NaX at 285 K temperature is 10 % dissociated. What will be the osmotic pressure? ($R=0.082$ L atm/mol /K)

• A. 5.2 atm
• B. 9.7 atm
• C. 12.9 atm
• D. 17.8 atm

Answer 1

The correct option is C 12.9 atm

$NaX\to N{a}^{+}+X^{-}$

Dissociation of 1 mole of NaX gives 2 moles of ions. So $n=2$

The degree of dissociation $\alpha =0.1$ as solute is 10% dissociated.

$\alpha =\frac{i-1}{n-1}$

$0.1=\frac{i-1}{2-1}$

The Van't Hoff factor

$i=1.1$

Now,

Osmotic pressure $\Pi =iCRT=1.1\times 0.5\times 0.082\times 285=12.9$ atm.