What is Molality and it's Formula?
Molality can be used for a measure of concentrations only when there is only one pure, clear solvent in the solution, but in certain cases, when the solution has more than one solvent in it, the problem arises. For example, in aqueous ethanol, when two solvents are present, either one could be treated as a solvent and the calculation, in that case, gets difficult. Not only this, in solid solutions like alloy, molality cannot be taken into calculation, and in that case, mole fraction is preferred over molality.
Molality is defined as the number of moles of solute present in 1000 gm of the solvent.
For example, 1 molal NaOH solution means a solution with 1 mole of NaOH in 1 Kg water.
Mathematically,
\[\text{molality(m)} = \frac{\text{number of moles of Solute(n)}}{\text{weight of the solvent in Kg}}\].
Here, molality is expressed as m and, mathematically, it is equal to moles of solute present per kilogram of the solution.
It is a basic SI unit of the amount of the substance which can be obtained by dividing the mass of the sample by the molar mass of the compound.
Mathematically,
\[\text{Number of moles} = \frac{\text{Mass of the given sample}}{\text{Molar mass of the element or compound}}\].
Molality does not depend on temperature. This is because mass is independent of temperature.
One key thing to note here is that we are using the weight of the ‘Solvent’ and not ‘Solution’ to calculate molality, which can be an easy mistake to commit while solving the problems.
FAQs on Molality Formula
1. What is the advantage of using molality over molarity?
Molality is preferred and has an added advantage in certain cases when the solution is highly concentrated and the solvent used is not water. Molarity depends upon temperature, so in cases where the temperature of the solution changes, the density also changes, and in that case, the molality of a solution is better than molarity. Hence, in these cases, it is better to use molality to express the concentration of the solution, especially because the mass of solute and mass of solvent does not depend upon temperature, and hence, they do not change. To determine the boiling point, melting point, or boiling point elevation and freezing point depression, molality is always a better option to be used.
2. What are the differences between molality and molarity?
Molality and Molarity, for a dilute aqueous solution, do not vary much, but for concentrated solutions, they vary. The main points of differences between them are as follows:
Molarity depends upon the mass of solvent in the solution, while molarity depends upon the volume of the solution.
Molality is expressed as m, while molarity is expressed as M.
Unit of molality is moles/Kg, while that of molarity is moles/litre.
Molality does not depend on the change of temperature of the solution, but molarity is affected by the change of temperature of the solution.
3. Calculate the molarity of a solution prepared from 20.20 grams of NaOH in 5.00 kg of water.
Solute = 20.20 gm of NaOH
Solvent = %.00 Kg of water
Molar mass of solute (NaOH) = 39.99 gm/mol
number of moles of solute = \[\frac{20.20}{39.99} = 0.5 moles \]
\[ Therefore, molality(m) = \frac{number of moles}{weight of solvent in Kg}\]
\[ \frac{0.5}{5}\]
\[m = 0.10\frac{moles}{kg}\]
4. A 6g sugar cube (Sucrose:C12H22O11 is dissolved in a 300 ml juice of 5° C water. What is the molality of the juice, given that density of water at 5° = 0.975 g/ml?
The weight of the Solute is 6 g of \[C_{12}H_{22}O_{11}\].
Molar mass of sucrose= 342 g/mol
Number of moles of solute, i.e., sucrose = 6g /(342 g/mol) = 0.0175 mol
Now,
Mass of water = density x volume
\[ = 0.975 * 300ml\]
\[ = 292.50 gm\]
\[ = 0.292 kg\]
Therefore,
Molality of the juice = 0.0175 mol / 0.292 kg molality = 0.059 mol/kg
Answer: The molality of the juice is 0.059 mol/kg.
