Question

A 0.004M solution of $$N{a}_{2}S{O}_4$$ is isotonic with a 0.010M solution of glucose at same temperature. The apparent degree of dissociation of $$N{a}_{2}S{O}_4$$ is:
(A) 25%
(B) 50%
(C) 75%
(D) 85%

Answer 1

Hint: The degree of association is the fraction of the total number of molecules of a substance which combine or associate to form bigger molecules. The formula of degree of association is $$\alpha =(i-1)$$. The term is usually used for the aggregation of molecules and atoms.
The degree of dissociation is defined as the generation of current carrying free ions, which have been dissociated from the fraction of solute at a given temperature. The formula for the degree of dissociation is $$\alpha ={\displaystyle \frac{i-1}{n-1}}$$.

Complete step by step solution:
As it is given in the question that the solutions are isotonic which means the osmotic pressure of sodium sulphate solution and the osmotic pressure of glucose solution is equal to each other. We know the formula of osmotic pressure and that is
Osmotic pressure= $$mRT$$
As the solution is isotonic so the osmotic pressure will be equal,
$${{\pi }_{N{a}_2}}_{S{O}_4}$$=$${\pi }_{glucose}$$
$$m_{1}RT\times i{}_1$$=$$m_{2}RT\times i_2$$
Now RT will get cancel and we will left with
$$m_1{i}_1$$=$$m_2{i}_2$$
$$\Rightarrow m_1$$=0.004M(molarity of sodium sulphate)
$$\Rightarrow i_2$$=1 (van't hoff factor of sodium sulphate)
$$\Rightarrow m_2$$= 0.010M (molarity of glucose)
$$\Rightarrow i_1$$=? ( van't hoff factor of glucose)
Now substituting the value in the above equation and we get
0.004 $$i_1$$=0.010 x 1
$$i_1$$= 2.5
The formula of degree of dissociation is
$$i=1+(n-1)\alpha$$
Here n is the number of ions in the compound.
When we will dissociate the compound we get
$$N{a}_{2}S{O}_{4}2Na+S{O}_4$$
The number of ions is 3.
So now substituting the values in degree of dissociation formula is
2.5=1+(3-1) $$\alpha$$
$$\Rightarrow \alpha$$=0.75
$$\Rightarrow \alpha$$=75%

Hence the correct answer for this question is option ‘C’.

Note: students generally forget to multiply the van't hoff factor with molarity of solution. Remember the units of molarity . Do show the dissociation equation for the ion. Understand the concept of association too. In solution dissociation works in dissociating the electrolyte or the ionic compound into the ions, so that the value of ‘n’ can be calculated. The dissociation property is different from dissociation constant. The formula for dissociation constant is $$K_a={\displaystyle \frac{{\alpha }^{2}C}{1-\alpha }}$$.