Molality may be a measure of the concentration of a solute during a solution in terms of the quantity of substance during a specified amount of mass of the solvent. Formation of term molality is in analogy to molarity which is that the molarity of a solution. G. N. Lewis and M. Randall within the 1923 publication of Thermodynamics and therefore the Free Energies of Chemical Substances was the earliest use of molality. During this chapter, we’ll discuss the molality formula, its advantages and solved examples.
The ratio of moles of solute to a kilogram of solvent is molality. Molality’s SI unit is moles per kilogram of solvent. The Systeme International d’Unites of units, the National Institute of Standards and Technology, US authority on measurement, considers the term “molal” and therefore the unit symbol “m” to be obsolete, and suggests mol/kg or a related unit of the SI. So, Molality= Moles of Solute/Kilogram of Solvent
In the case of solutions with over one solvent, molality for the mixed solvent considers a pure pseudo-solvent. Rather than mole solute per kilogram solvent as within the binary case, units are defined as mole solute per kilogram mixed solvent. Molality isn’t as common as its counterpart molarity but it’s utilized in very specific calculations, most notably involving colligative properties.
Advantages of Molality
Molality only depends on the masses of solute and solvent as it is the measure of concentration. And it is unaffected by variations in temperature and pressure. This is one of the advantages of molality. In contrast, solutions prepared volumetrically are likely to change as temperature and pressure change.
Another advantage of molality is that the proven fact that the molality of one solute during a solution is independent of the presence or absence of other solutes. Osmolality may be a variation of molality that takes under consideration only solutes that contribute to a solution’s pressure. It’s measured in osmoles of the solute per kilogram of water.
Solved Examples for Molality Formula
Q1] 5% (w/w) glucose solution is given .Find the molality of solution (density = 1.5gm/cc).
Solution: Molar mass of glucose = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol-1
5% (w/w) glucose solution is equal to 5 gm Glucose in 100 gm solution.
Therefore, Solvent is 95 gm
Now we have formula, Density = mass /volume
So volume = Mass/Density = 100/1.5 = 1000/15 ml
Formula for Molality (m) = W/MM × 1000/(W gm in solvent) = 5/180 ×1000/95 = 100/(18 ×19)=50/171 = 0.2923
Q2] A solution of glucose in water is labelled as 10% w/w, Find out the following using molality formula
- mole fraction of every component within the solution
Solution: 10% w/w solution of glucose in water means 10 g of glucose is present in 100 g of the solution i.e., 10 g of glucose is present in (100 – 10) g = 90 g of water.
Molar mass of glucose (C6 H12 O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g/mol
Then, number of moles of glucose = 10 / 180 mol = 0.056 mol
Therefore, Molality of solution = 0.056 mol / 0.09kg = 0.62 m
Number of moles of water = 90g / 18g mol-1 = 5 mol
Mole fraction of glucose (mg) = 0.056 / ( 0.056+5) = 0.011
And, mole fraction of water mw = 1 – mg = 1 – 0.011 = 0.989