To determine the van't Hoff factor (i) for the compound
X3Y that is
60% ionized in an aqueous medium, we need to consider the degree of ionization and the number of particles produced upon ionization.
For
X3Y, if it completely ionizes, it will dissociate as follows:
X3Y⟶3X++Y3−So, if completely ionized, one molecule of
X3Y produces a total of 4 ions (
3X+and
1Y3− ).
Given that the compound is only
60% ionized, the fraction of the compound that ionizes is 0.6 . Therefore, we need to determine the actual concentration of particles in the solution, taking into account both ionized and non-ionized forms.
Let's denote the initial number of molecules of
X3Y as 1 (for simplicity). When
60% ionizes, add the ion and the remaining non-ionized molecules:
Total particles
= (Number of non-ionized molecules
)+ (Number of ions from ionized molecules)
Total particles
=(1−0.6)+(0.6×4)Total particles
=0.4+2.4Total particles
=2.8 The van't Hoff factor (i) measures the effective number of particles in the solution and can be calculated as:
i=‌| ‌ Total particles in solution ‌ |
| ‌ Initial number of molecules ‌ |
For this case:
i=2.8Therefore, the van't Hoff factor for the compound
X3Y that is
60% ionized in aqueous medium is 2.8 .
The correct answer is:
Option D - 2.8