Question

A $ 0.2M $ sucrose solution is isotonic with $ 0.08M $ $ BaC{l_2} $ solution. Degree of ionization of $ BaC{l_2} $ is:
(A) 0.87
(B) 0.75
(C) 0.66
(D) 0.5

Answer 1

Hint :The cycle where water atoms accumulate around the polar covalent particle and because of fractional charges present on them and they attempt to pull separated the atom, so it is a cycle of creation of particles from particles as complete charges were absent upon them. Disintegration of HCl in water will be named as ionization.

Complete Step By Step Answer:
First let us split the $ BaC{l_2} $ into its consistent ions as:
 $ BaC{l_2} \to B{a^{2 + }} + 2C{l^ - } $
We were given that :
 $ {\Pi _1} = {C_1}RT = 0.2MRT $ …………..(1)
 $ {\Pi _2} = i{C_2}RT = i \times 0.08MRT $ …………....(2)
Now,
 $ {\Pi _1} = {\Pi _2} $
 $ \Rightarrow 0.2MRT = i \times 0.08MRT $
Let us simplify the above equation by cancelling the similar terms on the either side;
We get: $ i = \dfrac{{0.2}}{{0.08}} $
When we simplify the above, we get the value of I as:
 $ \Rightarrow i = \dfrac{{20}}{8} = 2.5 $
Now, let us take into consideration the equation we split in the first, that is:
 $ BaC{l_2} \to B{a^{2 + }} + 2C{l^ - } $
Let us calculate the ionization on each of the molecules and assign the value $ \alpha $ to the ones where we don’t know the ionization. So, as we calculate the charge, we find that the charge on
 $ BaC{l_2} $ is $ 1 - \alpha $
Charge on $ B{a^{2 + }} $ is $ \alpha $ and similarly,
Charge on $ 2C{l^ - } $ is $ 2\alpha $ .
Now, let us find out the value of ionization as:
 $ i = \dfrac{{(1 - \alpha ) + (\alpha ) + (2\alpha )}}{{1 + 0 + 0}} $
As we know the value of i let us substitute that value in the above expression,
 $ \Rightarrow 2.5 = \dfrac{{1 + 2\alpha }}{1} $ ,
 $ \Rightarrow 2.5 = 1 + 2\alpha $
Let us bring the value we need to find to the left hand side of the equation and the rest to the right hand side of the equation:
 $ \Rightarrow 2\alpha = 1.5 $
 $ \Rightarrow \alpha = \dfrac{{1.5}}{2} = 0.75 $
Therefore, we get the value of ionization $ \alpha $ as 0.75.
Thus, the correct answer to the above question is option B.

Additional Information:
In Ionic bond formation, when two oppositely charged particles approach one another ,the charged cation draws in electrons and repulses the positive core, this creates polarization in the ionic bond in view of that Ionic bonds create covalent character.

Note :
In sucrose, the monomers glucose and fructose are connected through an ether connection between C1 on the glucosyl subunit and C2 on the fructosyl unit. The bond is known as a glycosidic linkage. Glucose exists overwhelmingly as a combination of α and β "pyranose" isomers, however just the α structure connects to fructose. Fructose itself exists as a combination of α and β "furanose" isomers, yet just the β isomer connects to glucose.