Question

Vant hoff factor for a na2so4 solution which is 75% ionised is

Answer 1

Answer:

Na2so4----->2Na^+ +So4^2-at initial conc: Na2So4=1,2Na+=O,SO4^2-=O at final :Na2So4 =1-(alpha) i=1-(alpha)+2(alpha)+(alpha)I=1+2(alpha)given i=2

put the value of i in above equation ,you will get the value of Alpha

Answer 2

Answer:

The Van't Hoff factor for a Na₂SO₄ solution is equal to 2.5 when it is 75% ionized.

Explanation:

We have α which is degree of ionization.

The chemical equation for dissociation of Na₂SO₄:

                       $Na_{2}SO_{4}$   →  $2Na^{+}+SO^{2-}_{4}$

At time t            1 - α            2α          α  

Van't Hoff factor will be

         $i=i_{Na_{2}SO_{4}}+i_{Na^{+}}+i_{SO_{4}^{2-}}$

         $i=1-\alpha +2\alpha +\alpha \\ i=1+2\alpha$

Given that $\alpha =\frac{75}{100}=0.75$

Then  $i=1+2(0.75)=1+1.5$

   ∴             $i=2.5$

Hence the Van't Hoff factor would be equal to 2.5 for Na₂SO₄.