The molality, denoted by the lowercase letter m, it is a term used to describe the concentration of a solution. It can be defined as the number of moles of a solute dissolved in 1 kilogram of solvent.
It is equal to the moles of solute (the substance that dissolves) divided by the kilograms of solvent (the substance used to dissolve).
Where the moles of solute are given by the equation:
So in the end, the molality equation is
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Molality is a measure of the concentration of a solute in a solution. Used primarily when temperature is a concern.
Molality is not as common as its counterpart, molarity (moles of solute per liter of solvent), but it is used in very specific calculations, especially with regard to colligative properties (boiling point elevation, boiling point depression). of freezing).
Since volume is subject to variation due to temperature and pressure, molarity also varies according to temperature and pressure. In some cases, the use of weight is an advantage because the mass does not vary with environmental conditions..
It is important that the mass of solvent is used and not the mass of the solution. Solutions labeled with molal concentration are indicated with a lower case m. A 1.0 m solution contains 1 mole of solute per kilogram of solvent.
The preparation of a solution of a given molality is easy because it only requires a good scale. Both solvent and solute are mixed, rather than measured by volume.
The SI (International System) unit for molality is mol / kg, or moles solute per kg of solvent. A solution with a molality of 1 mol / kg is often described as "1 molal" or "1 m".
However, following the SI unit system, the National Institute of Standards and Technology, which is the United States authority on measurement, considers the term "molal" and the unit symbol "m" to be obsolete and suggests using mol / kg.
The preparation of a solution of a given molality involves weighing both the solute and the solvent and obtaining their masses..
But in the case of molarity, the volume of the solution is measured, which leaves room for density variations as a result of the environmental condition of temperature and pressure..
This means that it is advantageous to work with the molality that deals with the mass because in the chemical composition, the mass of a known pure substance is more important than its volume since the volumes can be altered under the effect of temperature and pressure while the mass remains unchanged.
Furthermore, chemical reactions take place in proportion to mass, not volume. To be based on mass, molality can easily be converted to a mass ratio or a mass fraction.
For aqueous solutions (solutions in which water is the solvent) near room temperature, the difference between molar and molar solutions is negligible..
This is because around room temperature, water has a density of 1 kg / L. This means that the "per L" of molarity is equal to the "per kg" of molality..
For a solvent such as ethanol, where the density is 0.789 kg / l, a 1 M solution would be 0.789 m.
The important part of remembering the difference is:
Molarity - M → moles per liter of solution.
Molality - m → moles per kilogram of solvent.
Problem: What is the molality of a solution containing 0.086 moles of NaCl dissolved in 25.0 g of water?
Answer: This example uses the direct molality equation:
Moles of solute (NaCl) = 0.086 mol
Mass (water) = 25.0 g = 25.0 x 10-3 kg
Substituting we have:
Problem: What is the molality of a solution if 124.2 grams of NaOH are dissolved in 1.00 liters of water? (The density of water is equal to 1.00 g / ml).
Answer: Water is known as the universal solvent and the density is usually given as 1.00 grams per milliliter, or what is the same as 1 kilogram per liter. This makes it easy to convert from liters to kilograms..
In this problem the 124.2 grams of NaOH will have to be converted into moles (for moles of solute) and one liter of water will have to be converted into kilograms (per kilograms of solvent) using the density.
First, find the molar mass of NaOH:
P.MNa= 22.99 g / mol
P.MOR= 16 g / mol
P.MH= 1g / mol
P.MNaOH= 22.99 g / mol +16 g / mol +1 g / mol = 39.99 g / mol
Second, you convert the grams of NaOH into moles, using the molar mass:
Third, convert the liter of water into kilograms using the density.
Solving for the mass remains
Finally the molality equation is solved:
Problem: What is the mass of water of an aqueous solution containing 0.52 grams of barium chloride (BaCltwo, PM = 208.23 g / mol) whose concentration is 0.005 mol / kg?
Answer: in this case the molal concentration is used to determine the amount of water in solution.
First, the number of moles of BaCl is obtainedtwo in the solution:
Second, remove kilograms of solvent from the molality equation and solve for.
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