Question

A solution contains 5.85g NaCl (Molar mass = 58.5 g/mol) per litre of
solution. It has an osmotic pressure of 4.75 atm at 27 C. Calculate the degree
of dissociation of NaCl in this solution.

Answer 1

Answer:

the degree

Explanation:

the clacyte of naci of dissociation of naci is solution

Answer 2

Answer:

The degree of dissociation of NaCl in this solution equals to 0.93.

Explanation:

The dissociation reaction:

           $NaCl$      →       $Na +Cl$

          (1 - α)                 α        α    

where  α is degree of dissociation

Total no. of moles $=1-\alpha +\alpha +\alpha =1+\alpha$

Van't hoff factor $i=\frac{1+\alpha }{1}$                                                  ..............(1)

Now the number of moles of NaCl $=\frac{5.85}{58.5}=0.1mol$

$V=1 L, T=273+27=298K$

Osmotic pressure,  $\pi =\frac{n}{V} RT$

$\pi =(0.1)(0.082LatmK^{-1}mol^{-1})(298K)\\\pi =2.46atm$

Observed osmotic pressure $=4.75atm$

Van't hoff factor, $i=\frac{\pi _{observed}}{\pi _{calculated}}$

Therefore $i=\frac{4.75}{2.46} =1.93$

Put the value of $i$ in eq.(1),

$1.93=1+\alpha \\\alpha =1.93-1\\ \alpha =0.93$

Therefore the degree of dissociation of NaCl comes out to be 0.93.