Definition

Molality of a given solution is defined as the total number of moles of solute per kilograms of solvent present in the solution. The molality of a solution is independent of the changes in physical properties of the system such as pressure and temperature as unlike volume, the mass of the system remains constant under all circumstances. Molality is represented by m, which is termed as molal. One molal as the molality of a solution where one gram of solute is dissolved in 1000 grams of solvent.

Molality Formula:

The equation for calculating molality is the ratio of the moles of solute whose molality is to be calculated and the amount of solvent used to dissolve the given solute.

$M= \frac{n}{W}$

Here, M is the molality of the solution that is to be calculated, n is the number of moles of the solute and W is the weight of solvent in Kgs.

Derivation:

Example 1:

For a solution prepared with 5 grams of toluene dissolved in 225 grams of benzene, calculate the molality.

Solution:

The chemical formula of Toluene = C7H8

The chemical formula for Benzene = C6H6

The molecular weight of Toluene = 12×7 + 1×8 = 92 moles/gram

The molecular weight of Benzene = 12×6 + 1×6 = 78 moles/gram

Given, mass of toluene in the solution = 5 g

The number of moles of Toluene =

$n_{T}= \frac{mass in grams}{molecular weight}$

$n_{T}= \frac{5}{92}= 0.054 Mole$

Now, given mass of Benzene = 225 g

Mass in terms of kg of the component =

$\frac{225 grams of Benzene}{1000}= 0.225 kilograms$

Now, we calculate the molality of the solution using the formula given above.

$molality = \frac{Number of moles of the element}{Mass in kg of the solvent}$

$Molality= \frac{0.054}{0.225}= 0.24 m$

The molality of the solution is 0.24m.

Example 2:

A solution prepared using 15 g of sodium sulphate dissolved in 125 ml of water at a temperature of 25ᵒC. The density of water at 25ᵒC is 1 g/cm3. Calculate the molality of the given solution of sodium sulphate.

Solution:

The molecular formula for sodium sulphate is Na2SO4.

The molecular formula for water is H2O.

The molecular mass of sodium sulphate is calculated as given below,

M=23×2+32+16×4=142

The number of moles of sodium sulphate in the given question is calculated as,

$n= \frac{mass in grams}{molecular weight}= \frac{15}{142}= 0.106$

Given, volume of solvent taken at 25ᵒC is 125 ml.

Expressing the amount of solvent in terms of mass,

m=d×V

Here, d is the density of the solvent at the given temperature and V is the volume.

m=1×125=125 g

Expressing the mass of solvent in terms of Kgs,

$m= \frac{125}{1000}= 0.125kg$

Now, using the formula given above, we calculate the molality of the given solution.

$molality= \frac{number of moles of solute}{mass of solvent in Kgs}$

Substituting the values, we get,

$molality= \frac{0.106}{0.125}$

The molality of the given solution is 0.85m.