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# Molality Formula

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.

"Molality is defined as the number of moles of solute dissolved in 1kg (1000g) of solvent."

Molality Formula is given by

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## Molality Problems

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 H2O?
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 H2O $\times$ 1kg / 1000g = 0.0501 kg H2O
molality = 0.0110 mol AA/0.0501 kg H2O
molality = 0.220m in AA

Question 2: A solution is prepared at 25oC by mixing 15g of Na2SO4 with 125cm3 of H2O. The density of H2O at this temperature is 1.0g/cm3. What is the molality of Na2SO4 in the solution?
Solution:

Molality is defined as the amount of solute dissolved in 1kg of solvent.
There is 15g Na2SO4 $\times$ (1 mol Na2SO4/142g Na2SO4) = 0.106mol of Na2SO4 and there is 125cm3 $\times$ 1.0g/cm3 = 125g of H2O.

Substituting into the expression for molality we have,

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

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