What is the relationship between depression in freezing point of a solution and molecular mass of the solution?

Answer

Verified

Hint: By using the formula of depression in freezing point one can determine the relation between the freezing point and the molecular weight. The depression in freezing point is equal to the product of cryoscopic constant and molality.

Complete step by step answer:

The freezing point depression is explained as the lowering of the freezing point of the solvent with the addition of the solute.The freezing point depression is the colligative property which is proportional to the molality of the solute added.The freezing point depression is given by the formula as shown below.$\Delta {T_f} = {K_f} \times m$Where,$\Delta {T_f}$ is the freezing point depression${K_f}$ is the cryoscopic constantm is the molalityThe molality is defined as the number of moles of solutes dissolved in one kilogram of solvent.The formula of molality is shown below.$m = \dfrac{n}{{{M_1}}}$Where,m is the molalityn is the number of moles${M_1}$ is the mass in kilogramThe number of moles is given by the formula as shown below.$n = \dfrac{m}{{{M_2}}}$Where,n is the number of molesm is the mass${M_2}$ is the molecular weightSo, if we substitute the terms in the formula of freezing point depression then the new formula obtained will be.$ \Rightarrow \Delta {T_f} = {K_f} \times \dfrac{m}{{{M_2}{M_1}}}$If we keep the ${K_f}$, m and ${M_1}$ as constant then the freezing point depression will be inversely proportional to the molecular weight.$ \Rightarrow \Delta {T_f} = \dfrac{1}{{{M_2}}}$So, by increasing the value of depression in freezing point, the molecular weight decreases and vice versa.

Thus, an increase in the molecular weight will have a smaller effect on the freezing point.

Note:

The freezing point is calculated by subtracting the depression in freezing point value and the freezing point of pure water which is zero degree Celsius.

1. The freezing point depression (ΔTf) is directly proportional to the molality of solution.

Thus, ΔTf = Kf m ….(1)

2. Suppose we prepare a solution by dissolving W2 g of solute in W1 g of solvent.

Moles of solute in W1 g of solvent = `"W"_2/"M"_2`

where, M2 is the molar mass of solute.

Mass of solvent = W1 g = `("W"_1 g)/(100 "g"//"kg") = "W"_1/1000` kg

3. The molality is expressed as,

m = `"Moles of solute"/"Mass of solvent in kg"`

m = `("W"_2//"M"_2 "mol")/("W"_1//1000 "kg")`

m = `(1000 "W"_2)/("M"_2 "W"_1) "mol kg"^-1` ....(2)

4. Substituting equation (2) in equation (1), we get,

`triangle "T"_"f" = (1000 "K"_"f" "W"_2)/("M"_2 "W"_1)`