Question

Calculate the molality of $12.5\%$ $w/w$ sulphuric acid?

Answer 1

In order to answer let us first know about molality. Molality is the number of moles of solute in a solution that corresponds to $1$ kilogram or $1000$ gram of solvent. The definition of molarity, on the other hand, is based on a certain volume of solution. Molality is measured in mol/kg, which is a standard unit in chemistry.

Complete answer:
Molality can also be referred to as molal concentration. It is a metric for determining the concentration of solutes in a solution. The solution is made up of two parts: the solute and the solvent. The concentration of a solution can be expressed in a variety of ways, including molarity, molality, normalcy, formality, volume percentage, weight percentage, and part per million. The mass of the solvent and the moles of solute must be calculated in this phrase.
Formula of Molality:-
$Molality\left( M \right) = \dfrac{{Number\,of\,moles\,of\,solute}}{{Mass\,of\,solvent\,in\,kgs}}$
$Molality\left( M \right) = \dfrac{{g \times 1000}}{{W \times m}}$
Now, coming to the asked question-
In a $100{\text{ }}g$ of solution, $12.5\%$ $\dfrac{w}{w}$ is $12.5{\text{ }}g$
Weight of the solvent = $100g - 12.5g = 87.5g$
The Number of moles of sulphuric acid is $\left( {{H_2}S{O_4}} \right)$ = $\dfrac{{12.5}}{{98}}$
Therefore, Number $\left( {{H_2}S{O_4}} \right)$ of moles of = $0.127{\text{ }}mol$
$Molality\left( M \right) = \dfrac{{0.127 \times 1000}}{{87.5}}$
$Molality\left( M \right) = 1.45m$
Therefore the molality of the $12.5\%$ $w/w$ sulphuric acid is $1.45m$ .

Additional information:
The main advantage of employing molality as a concentration measure is that it is based solely on the masses of the solute and solvent, which are unaffected by temperature and pressure changes. Volumetric solutions (e.g. molar concentration or mass concentration) are less likely to vary as temperature and pressure change. This is a considerable advantage in many applications since the mass, or amount, of a substance is often more essential than its volume (e.g. in a limiting reagent problem). Another benefit of molality is that one solute's molality in a solution is unaffected by the presence or absence of other solutes.

Note:
One point should be noted about molality is that when there is no pure ingredient present in a mixture, molality does not apply. Mixtures of water and alcohol, for example, or alloys. Any chemical can be used as a solvent in this situation.