Question

Calculate the amount of \[CaC{l_2}\] (molar mass = 111 g/mol) which must be added to 580 g of water so that freezing point lower by 2 K, assuming \[CaC{l_2}\] is completely dissociated:
[ \[{K_f}\] for water = 1.86 K kg/mol]

Answer 1

Hint: To solve this question, we must first calculate the molality of the given solute. Then we must calculate the Van’t Hoff factor and then substitute all these values in the formula for calculating the depression of the freezing point to find the value of the amount of solute used.

Formula used: \[\Delta {T_f} = i.{K_f}.m\]

Complete Step-by-Step Answer:
Before we move forwards with the solution of the given question, let us first understand some important basic concepts.
The number of moles of a substance can be calculated as the ratio of the weight of the sample given to us, to the molecular weight of the given element or compound.
Molality can be calculated as the ratio of the number of moles of the solute to the weight of the solvent used.
Also, the number of particles of the solute also matters when we calculate the depression in freezing point. In case of covalent solutes, the number of particles remains constant. Whereas when the solute di – ionic in nature, then due to dissociation, the number of particles becomes two times the original consideration. To overcome this error, we introduce a correction factor known as Van’t Hoff factor. Its value is equivalent to the number of ions the given solute dissociates into.
Hence, the depression in the freezing point can be calculated as:
 \[\Delta {T_f} = i.{K_f}.m\]
Where, \[\Delta {T_f}\] is the depression in the freezing point, i is the Van’t Hoff factor, m is the molality of the solute, \[{K_f}\] is the cryogenic constant.
Hence,
 \[\Delta {T_f} = i.{K_f}.(\dfrac{n}{W})\]
n \[ = \dfrac{{weight\,\,of\,\,solute}}{{molar\,\,mass\,\,of\,\,solute}} = \dfrac{x}{{111}}\]
Also, \[CaC{l_2}\] dissociates into 3 ions. Hence, \[i = 3\] .
 \[\Delta {T_f} = i.{K_f}.(\dfrac{n}{W})\]
 \[2 = (3).(1.86).(\dfrac{x}{{(580)(111)}})\]
x = 23.07 g

Hence, the amount of \[CaC{l_2}\] which must be added to 580 g of water so that freezing
point lower by 2 K, assuming \[CaC{l_2}\] is completely dissociated is 23.07 g

Note: Another thing we must understand is the depression of the freezing point of a given pure liquid. When we add any other substance to a sample of a pure liquid, it becomes more difficult for the liquid to lose energy. Because of this, we must reduce the temperature of the system for the liquid to freeze. This lowers the freezing point of the liquid.