Molality it is a form of concentration of solutions (like common concentration, concentration in mol/L, mass title), used to determine the relationship between the amount in mol of solute in a given mass of solvent.
The symbol used to represent the molality is the W; O mole number of the solute is represented by n1; and the mass of the solvent is symbolized by m2. Thus, with these symbols, we can build the mathematical expression using the proposed definition for molality:
W = no1
m2
This relationship is expressed mathematically by a division, therefore, the molality is a form of concentration that involves dividing the number of moles of the solute by the mass of the solvent.
It is noteworthy that the number of moles of the solute has a specific mathematical expression, since it is the relationship between the mass of the solute (m1), contained in the solution, and the molar mass of this solute (M1):
no1 = m1
M1
So, from that, we can associate the formula for the number of moles of the solute with the formula for molality, resulting in the following expression:
W = m1
M1.m2
Regardless of the formula used, whether associated with the mol number formula or not, the unit of measure used must be mol/Kg or molal. Thus, it is important to observe the data with the following units:
Solute mass (m1): grams (g)
Molar mass of solute (M1): grams per mol (g/mol)
Number of moles of the solute (n1): mol
Solvent mass (m2): kilogram (Kg).
Below are some examples of exercises on the molality:
1st Example - (ITA-SP) The label on a bottle says it contains a 1.50 molal LiNO solution3 in ethanol. This means that the solution contains:
a) 1.50 mol of LiNO3/kilogram of solution.
b) 1.50 mol of LiNO3/liter of solution.
c) 1.50 mol of LiNO3/kilogram of ethanol.
d) 1.50 mol of LiNO3/liter of ethanol.
e) 1.50 mol of LiNO3/mol of ethanol.
In this exercise, we should only perform a theoretical analysis, as it reports the molality (1.5 molal of LiNO3 in ethanol) and asks for the composition of the solution. We know that molality is the relationship between the mol number of the solute (LiNO3) and the mass of the solvent (ethanol), and that the term molal can be replaced by mol/Kg. Therefore, we have this in the solution (alternative c):
Do not stop now... There's more after the advertising ;)
1.5 mol of LiNO3;
1 kg of ethanol.
2nd Example - (UFF-RJ-Adapted) Glucose, with structural formula C6H12O6, is a simple sugar and is also the main source of energy for humans and other vertebrates. More complex sugars can be converted to glucose. In a series of reactions, glucose combines with the oxygen we breathe and produces, after many intermediate compounds, carbon dioxide and water with the release of energy. Hospital intravenous feeding usually consists of a solution of glucose in water with the addition of mineral salts. Assuming 1.50 g of glucose are dissolved in 64.0 g of water, what will be the molality of the resulting solution?
a) 0.13
b) 0.20
c) 0.40
d) 0.31
e) 0.41
The data provided by the exercise are:
Solute mass (m1): 1.5 g
Solvent mass (m2): 64 g
Molecular formula of solute: C6H12O6
To determine the molality of the solution, it is interesting to perform the following steps:
1st Step: Transform solvent mass from g to kg.
To do this, just divide the supplied mass, 64 g, by 1000, which results in 0.064 kg.
2nd Step: Determine the molar mass of the solute.
For this, we must multiply the number of atoms of the element in the molecular formula by its atomic mass (present in the periodic table) and then add the results:
M1 = 6.mass of C + 12. mass of H + 6. mass of O
M1= 6.12 + 12.1 + 6.16
M1 = 72 + 12 + 96
M1 = 180 g/mol
3rd Step: Use the data provided and found in the previous steps in the following expression:
W = m1
M1.m2
W = 1,5
180.0,064
W = 1,5
11,52
W = 0.13 mol (approximately)
By Me. Diogo Lopes Dias
