Question
Question: A 0.0020M aqueous solution of an ionic compound \(Co{{(N{{H}_{3}})}_{5}}(N{{O}_{2}})Cl\)freezes at -...
A 0.0020M aqueous solution of an ionic compound Co(NH3)5(NO2)Clfreezes at -0.00732°C. Number of moles of ions which 1 mole of ionic compound produces on being dissolved in water will be: [Kf=1.86∘C/m]
(a) 1
(b) 2
(c) 3
(d) 4
Solution
Van’t Hoff factor is the number of splitting or forming of ions of the solute in a solution. To solve this, you can use the relation of the Van’t Hoff factor with the dissociation constant and find out the correct answer.
For dissociation, i=1+α(n−1) and for association, i=1−(1−n1)α.
Also, for calculating depression in freezing point (ΔTf) will be calculated using the formula: ΔTf=i×Kf×m
Complete step by step solution:
We have been provided with an aqueous solution of an ionic compound Co(NH3)5(NO2)Cl with concentration: 0.0020M, and freezing point: -0.00732°C.
Now, depression in freezing point (ΔTf) is given by: ΔTf=i×Kf×m,
Where, i=Vant Hoff factor
Kf= cryoscopic constant (for water: 1.86°C/m),
m= molality of solution
In this question, we have been given ΔTf=0−(−0.00732),
So, depression in freezing point comes out to be: ΔTf=0.00732,
Now, as we need to calculate number of moles of ions, so for that we will be calculating Van't Hoff factor(i) using the formula:i=Kf×mΔTf,
Keeping the values, we would get: i=1.86×0.00200.00732,
So, Van't Hoff Factor comes out to be: i=1.86≈2,
So, the compound Co(NH3)5(NO2)Cl in solution will dissociate into [Co(NH3)5(NO2)]+ and Cl−,
So, the number of moles of ions that is n comes out to be: n=2.
Therefore, option (b) is correct.
Note: Non-electrolytes dissolved in water have the value of i equal to 1. Ionic compounds dissolve in water and thus the Van’t Hoff factor is the number of ions in a formula unit of the substance.
Van't Hoff Factor for dissociation is: i=1+α(n−1).