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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)ClCo{{(N{{H}_{3}})}_{5}}(N{{O}_{2}})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=1.86C/m]\left[ {{K}_{f}}=1.86{}^\circ C/m \right]
(a) 1
(b) 2
(c) 3
(d) 4

Explanation

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(n1)i=1+\alpha \left( n-1 \right) and for association, i=1(11n)αi=1-\left( 1-\dfrac{1}{n} \right)\alpha .
Also, for calculating depression in freezing point (ΔTf)\left( \Delta {{T}_{f}} \right) will be calculated using the formula: ΔTf=i×Kf×m\Delta {{T}_{f}}=i\times {{K}_{f}}\times m

Complete step by step solution:
We have been provided with an aqueous solution of an ionic compound Co(NH3)5(NO2)ClCo{{(N{{H}_{3}})}_{5}}(N{{O}_{2}})Cl with concentration: 0.0020M, and freezing point: -0.00732°C.
Now, depression in freezing point (ΔTf)\left( \Delta {{T}_{f}} \right) is given by: ΔTf=i×Kf×m\Delta {{T}_{f}}=i\times {{K}_{f}}\times m,
Where, i=i=Vant Hoff factor
Kf={{K}_{f}}= cryoscopic constant (for water: 1.86°C/m),
m= molality of solution
In this question, we have been given ΔTf=0(0.00732)\Delta {{T}_{f}}=0-(-0.00732),
So, depression in freezing point comes out to be: ΔTf=0.00732\Delta {{T}_{f}}=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=ΔTfKf×mi=\dfrac{\Delta {{T}_{f}}}{{{K}_{f}}\times m},
Keeping the values, we would get: i=0.007321.86×0.0020i=\dfrac{0.00732}{1.86\times 0.0020},
So, Van't Hoff Factor comes out to be: i=1.862i=1.86\approx 2,
So, the compound Co(NH3)5(NO2)ClCo{{(N{{H}_{3}})}_{5}}(N{{O}_{2}})Cl in solution will dissociate into [Co(NH3)5(NO2)]+\left[ C \right.o{{(N{{H}_{3}})}_{5}}(N{{O}_{2}}{{\left. ) \right]}^{+}} and ClC{{l}^{-}},
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(n1)i=1+\alpha (n-1).