Calculation of Molality of a Sparing Soluble Salt Solution

Given:
Vapour pressure of a very diluted saturated solution of a sparingly soluble salt X3Y4 = 20 mm Hg
Vapour pressure of pure water = 20.0126 mm Hg

To calculate:
Molality (m) at temperature T

Solution:

Determine the Vapour Pressure Lowering of the Solution

The vapour pressure of a solution is lower than that of the pure solvent due to the presence of solute particles. The lowering in vapour pressure is directly proportional to the mole fraction of the solute. Therefore, we can use Raoult's law to calculate the mole fraction of the solute.

Ptotal = P1 * X1 + P2 * X2
where,
Ptotal = total vapour pressure
P1 = vapour pressure of pure solvent
X1 = mole fraction of solvent
P2 = vapour pressure of solute
X2 = mole fraction of solute

In this case, the vapour pressure of pure water is 20.0126 mm Hg and the vapour pressure of the solution is 20 mm Hg. Therefore, the vapour pressure lowering of the solution is:

ΔP = P1 - Ptotal
ΔP = 20.0126 - 20
ΔP = 0.0126 mm Hg

Calculate the Molality of the Solution

The molality (m) of a solution is defined as the number of moles of solute per kilogram of solvent. We can use the following formula to calculate the molality of the solution:

ΔP = Kf * m * i
where,
Kf = freezing point depression constant
m = molality of the solution
i = van't Hoff factor (number of particles per molecule)
ΔP = vapour pressure lowering of the solution

For a very diluted solution, we can assume that the molality of the solute is equal to its molarity (M). Therefore,

M = moles of solute / kg of solvent

Since the solution is very diluted, we can assume that the mass of the solvent is equal to the total mass of the solution. Therefore,

kg of solvent = kg of solution

Now we can substitute these values in the formula to get:

ΔP = Kf * M * i

Solving for M, we get:

M = ΔP / (Kf * i)

Determine the Freezing Point Depression Constant (Kf)

The freezing point depression constant (Kf) is a constant that is specific to a particular solvent. For water, the value of Kf is 1.86 °C/m. Since we are working with vapour pressure lowering, we need to convert this value to units of mm Hg/m.

Kf = 1.86 °C/m = 1.86 K/m = 1.86 * 1000 °C/m = 1860 °C/m
ΔTf = Kf * m
where,
ΔTf = freezing point depression

Since we are not given the freezing point depression, we can assume that the temperature is close to the normal boiling point of water (100 °C). Therefore, we can use the following formula to convert the freezing point depression in units of Celsius to units of