Answer: Option (a) is the correct answer.
Explanation:
Depression in freezing point depends on Vant Hoff factor. So, more is the number of dissociated ions lesser will be the freezing point.
Mathematically, [tex]\Delta T_{f} = iK_{f}m[/tex]
where [tex]\Delta T_{f}[/tex] = freezing point depression
i = Vant Hoff factor
[tex]K_{f}[/tex] = freezing point depression
m = molality of the solution
The ratio between concentration of particles that are actually produced when a substance is dissolved and the concentration of a substance calculated from its mass is known as Vant Hoff factor.
Mathematically, i = [tex]\frac{\text{total no. of moles of particles after association/dissociation}}{\text{No. of moles of particles before association/dissociation}}[/tex]
For dissociation i is greater than or equal to 1. 1 then dissociation has taken place. When i = 1 then neither association or dissociation has taken place.
Hence, sucrose will have i = 1 as it produces only one ion.
Sodium chloride will have i = 2 as it produces two ions.
Calcium bromide will have i = 3 as it produces three ions.
Aluminium bromide will have i = 4 as it produces four ions.
Potassium chloride will have i = 2 as it produces two ions.
Therefore, we can conclude that sucrose solution will have highest freezing point.
