Molality abbreviated mol/kg or m, specifies the number of moles of solute per kilogram of solvent. The primary advantage of using molality to specify concentration is that unlike its volume, the mass of the solvent does not change with changes of temperature or pressure. So molality remains constant under changing environmental conditions.

Molality is another way of expressing the concentration of a solution. It can be defined as the number of moles of a solute dissolved in 1 kilogram of the solvent. The symbol for molality is m (small m in italics) or –m (small m with a hyphen). This is to differentiate molality from molarity which is designated by capital letter M; and to different it from mass, which is symbolized with small letter m.

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Solved problems based on molality are given below. ### Solved Examples

**Question 1: **Find the molality of ascorbic acid (MW = 176) in a solution prepared from 1.94g of ascorbic acid and 50.1g of H_{2}O?

** Solution: **

In order to find the molality (m) we need to express the amount of solute in moles and the mass of solvent in kg.

1.94g AA $\times$ 1mol AA/ 176g AA = 0.0110 mol AA

50.1g H_{2}O $\times$ 1kg / 1000g = 0.0501 kg H_{2}O

molality = 0.0110 mol AA/0.0501 kg H_{2}O

molality = 0.220m in AA

**Question 2: **A solution is prepared at 25^{o}C by mixing 15g of Na_{2}SO_{4} with 125cm^{3} of H_{2}O. The density of H_{2}O at this temperature is 1.0g/cm^{3}. What is the molality of Na_{2}SO_{4} in the solution?

** Solution: **

Molality is defined as the amount of solute dissolved in 1kg of solvent.

There is 15g Na_{2}SO_{4} $\times$ (1 mol Na_{2}SO_{4}/142g Na_{2}SO_{4}) = 0.106mol of Na_{2}SO_{4} and there is 125cm^{3} $\times$ 1.0g/cm3 = 125g of H_{2}O.

Substituting into the expression for molality we have,

Molality = $\frac{0.106\ mol}{125g}$ $\times$ $\frac{1000g}{1kg}$ = 0.85m

In order to find the molality (m) we need to express the amount of solute in moles and the mass of solvent in kg.

1.94g AA $\times$ 1mol AA/ 176g AA = 0.0110 mol AA

50.1g H

molality = 0.0110 mol AA/0.0501 kg H

molality = 0.220m in AA

Molality is defined as the amount of solute dissolved in 1kg of solvent.

There is 15g Na

Substituting into the expression for molality we have,

Molality = $\frac{0.106\ mol}{125g}$ $\times$ $\frac{1000g}{1kg}$ = 0.85m