Molality is used to indicate the concentration of a given solution and is also known as molal concentration. Molality is represented as a number of moles of solute in proportion to the mass of the solvent in the solution. And the mass of solvent is taken in kilogram units, so the commonly used unit for molality is mol/kg of solvent.
Generally, we calculate concentration with reference to the volume of the solution referred to as molarity. Molarity, also termed as molar concentration, is the number of moles of solute in proportion to the volume of solution. The first advantage is that molality always remains constant for a given solution, but molarity can alter. The second advantage is that the molality of one solute is independent of the other solutes because it is the mass of solvent, not solution.
The change in pressure or temperature cannot affect the mass but can change the volume of a solution. So, molality can be very beneficial to compare and determine the colligative properties of a given solution. Molality is best suited for calculating these colligative properties of the solution, such as boiling point, melting point, boiling point elevation, freezing point depression. We will see the difference while calculating molality in the examples below.
What is Molality
The molality of any given solution is defined as the measure of the number of moles of the solute present in 1 kilogram of solvent.
Molality Formula
We determine the molality of the solution by dividing the number of moles present in the solute by the mass of solvent.
The unit of molality is moles of solute per kilogram of solvent and is represented by italicized and small lettered ‘m’ or ‘molal’. SI unit of molality is mol/kg. It means a solution of concentration 1m or 1molal is having a molality of 1 mol/kg. Molality should not be confused with molarity which is represented by the capital letter ‘M’ or ‘molar’ and is denoted by moles/liter.
How To Calculate Molality
Calculations for molality are broadly outlined in the following steps:
- Firstly write down the equation and determine the solute and solvent present in the given solution.
- Calculate the total number of moles present in the solute.
- Then calculate the mass of the solvent in a kilogram unit.
- Finally, calculate the molality by dividing the amount of solute (calculated in Step 2) with the mass of solvent (calculated in Step 3).
Example 1:
Let’s say if a chemist dissolves 7 grams of magnesium chloride in 500 ml of water. What will be the molality of the given solution?
Solution
First step:
Firstly we need the chemical formula for Magnesium chloride, which is MgCl2. Now we have to calculate the molar mass of this compound. For calculating the molar mass of a compound with multiple atoms, we will sum up the atomic masses of the individual atoms.
In this particular example, atomic masses of Magnesium and Chlorine are 24.31 g/mol and 35.45 g/mol, respectively. So, the molar mass will be 95.21 g/mol.
Molar mass of MgCl2 = 1(Atomic mass of Mg) + 2(Atomic mass of Cl)
= 1(24.31 g/mol) + 2(35.45 g/mol)
= 95.21 g/mol
Second step:
Next, we will convert the given mass to moles of solute present in the solution. If 95.21 g makes up 1 mole, so 7 g will contain 0.0735 moles.
95.21 g has moles = 1 mole
So, 7 g has moles = 7 g X (1 mol) / (95.21 g/mol) = 0.0735 moles
Third step:
Let us calculate the mass of solvent, which is water, in this particular example. Generally, the mass and volume of water always remain the same as the density of water is 1 gram per milliliter (unless assumptions are provided with the question). So, here the volume of solvent is 500 ml. So using the formula Density = Mass / Volume, we will get the mass of water which will be 500 gm of mass. The solvent mass has to be in kilograms, so we will convert grams to kilograms, which equals 0.5 kilograms.
Fourth step:
Now, we can calculate molality using the earlier mentioned formula:
Molality = moles of solute / kg of solvent
= 0.0735 moles / 0.5 kg = 0.147 mol/kg
So, Molality of the solution is 0.147 mol/kg.
Example 2:
If you have 700 ml of hot water, having a temperature of 80°C in a glass, and you dissolve 8 gram sugar cube in it. What will be the molality of the final sugar solution? Assume: Water density at 80°C = 0.975 g/ml
Solution
First step:
A sugar cube is made of sucrose having the chemical formula: C12H22O11
We will start with determining the number of moles of sucrose in 8 grams. For this, we will need atomic masses of all the three atoms carbon, hydrogen, and oxygen, which are 12.01 g/mol, 1.008 g/mol, and 16.0 g/mol, respectively
Molar mass C12H22O11 = 12(Atomic mass of C) + 22(Atomic mass of H) + 11(Atomic mass of O)
= 12(12.01) + 22(1.008) + 11(16.00)
= 342.296 g/mol
Second step:
Now, we will calculate moles in solute from the molar mass. If 342.296 g makes up 1 mole, so 8 g will contain 0.0233 moles.
342.296 g has moles = 1 mole
So, 8 g has moles equal to = 8 g X (1 mol) / (342.296 g/mol) = 0.0233 moles
Third Step:
Next, we will calculate the mass of the water in 700 ml, which acts as a solvent in a given solution. In the previous example, this conversion was not required. But in this one, we will be using the formula Density = Mass/Volume as there is a change in density of water at a hot temperature of 80°C.
Density = Mass/Volume
which means Mass = Density x Volume
= 0.975 g/ml x 700 ml
= 682.5 g
As we use the mass of solvent in kilogram for calculating molality, this equals 0.682 kg.
Fourth step:
Finally, using the above calculations, we can easily calculate the molality of the given sugar solution.
Molality = moles of solute / kg of solvent
= 0.0233 moles / 0.682 kg = 0.0341 mol/kg
So, Molality of the sugar solution is 0.0341 mol/kg.
Tip:
Since the density of dilute aqueous solutions is comparatively close to 1 gram per milliliter, molality and molarity are closely related. This indicates that 1 liter of the solution has a mass of 1 kilogram. But, the density of the solution will not remain equal to 1 gram per milliliter as it becomes more condensed, which will result in variation in molality and molarity.
Similarly, the molality of solutions with solvents other than water would be very different from the molarity. So for those calculations, we need to pay attention while working with the density formula mentioned above.
