Question

0.067 molar aqueous solution of a binary electrolyte $ {A^ + }{B^ - } $ shows $ 2.46{\text{ atm}} $ osmotic pressure at $ {27^\circ }{\text{C}} $ . What fraction of $ {A^ + }{B^ - } $ remains unionised?
(A) $ 10\% $
(B) $ 15\% $
(C) $ 50\% $
(D) Zero

Answer 1

Hint: In Ionic equilibrium, the ionic substance in polar solvents dissociates into their ions. The ions formed in the solution are always in equilibrium with their undissociated solution. In their aqueous solution, electrolytes dissociate into their component ions and thus conduct electricity. We shall use the colligative property of osmotic pressure to solve this question.

Formula Used:
We will use the following formula to find out the required solution if the given question:
 $ \pi = i \times C \times R \times T $
Which can be re written as
 $ i = \dfrac{\pi }{{CRT}} $
Where
 $ i $ is the van't Hoff factor
 $ \pi $ is the osmotic pressure
 $ T $ is the temperature in kelvin
 $ C $ is the concentration expressed as molarity
 $ R $ is the gas constant whose value is $ 0.0821 $ .

Complete Step-by-Step solution:
According to the question, we have been provided with an electrolyte $ {A^ + }{B^ - } $ .
The concentration of a binary electrolyte is $ C = 0.067{\text{ M}} $
The osmotic pressure is given as $ \pi = 2.46 {\text{atm}} $
The temperature is given as $ T = {27^\circ }{\text{C}} = 300 {\text{K}} $
The given electrolyte can be broken down as
 $ AB \rightleftharpoons {A^ + } + {{\rm B}^ - } $
Now, let us suppose the concentration of $ AB $ as $ 1 $
Then, the concentration of $ {A^ + } $ and $ {{\rm B}^ - } $ will be nothing
Similarly,
If the concentration of $ AB $ is $ 1 - \alpha $ , then the concentration of both $ {A^ + } $ and $ {{\rm B}^ - } $ will be $ \alpha $ .
Let us understand this with the help of easy representation
 $ AB \rightleftharpoons A { ^ + } + {{\rm B}^ - } $
 $ 1 - - $
 $ 1 - \alpha \alpha \alpha $
Now, the total number of particles will be
 $ 1 - \alpha + \alpha + \alpha = 1 + \alpha $
We know that $ 1 + \alpha = i $
Now, let us use the formula and substitute the values
 $ i = \dfrac{\pi }{{CRT}} $
 $ = \dfrac{{2.46}}{{0.067 \times 0.0821 \times 300}} $
Upon solving, we get
 $ i = 1.5 $
Now, $ 1 + \alpha = i $
Let us substitute the value of $ i = 1.5 $
 $ 1 + \alpha = 1.5 $
So, we get
 $ \therefore \alpha = 0.5 $ that is $ 50\% $
Hence, the correct option is (C).

Note:
The minimum pressure to be applied to a solution to stop the fluid motion of solvent molecules through a semipermeable membrane can be defined as osmotic pressure (osmosis). It is a colligative property and is reliant on the solute particle concentration in the solution.