On a besoin de préparer 15,0 litres d'une solution de HCl 0,5 M. On dispose pour cela d'un acide

step 1 All right, here's the problem 60. You can purchase hydrochloric acid in a concentrated form that

On a besoin de préparer 15,0 litres d'une solution de HCl...

On a besoin de préparer 15,0 litres d'une solution de $\mathrm{HCl} 0,5 \mathrm{M}$. On dispose pour cela d'un acide chlorhydrique commercial, qui est une solution à $36,6 \%$ de $\mathrm{HCl}$ (en $\mathrm{g}$ pour $100 \mathrm{~g}$ de solution), et dont la masse volumique est $1,18 \mathrm{~kg} \cdot \mathrm{L}^{-1}$. Quel volume de solution commerciale faut-il prendre, et quel volume d'eau faut-il y ajouter?

Key Concepts

Molarity

Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. It is used to relate the amount of substance in a given volume solution to the required chemical reaction or process. In solution preparation and dilution, molarity helps determine how much solute is present and how much solvent is needed to reach a desired concentration.

Mass Percentage

Mass percentage, often represented as a percentage concentration, describes the ratio of the mass of a solute to the total mass of the solution, multiplied by 100. This concept is key when working with commercial solutions that are specified by their weight percentage, as it allows conversion from mass-based concentration into the molarity needed for precise volumetric calculations.

Density

Density is the ratio of the mass of a substance to its volume and is typically expressed in units such as kg/L for solutions. In the context of solution preparation, density is essential for converting between mass and volume, particularly when commercial solutions are provided with a given density, thereby allowing accurate calculation of the amount of solute available in a specific volume.

Dilution

Dilution involves reducing the concentration of a solution, usually by adding more solvent. It is governed by the principle that the number of moles of solute remains constant before and after dilution, which is reflected in the equation M1V1 = M2V2. Understanding dilution is crucial for preparing solutions of a desired concentration from a stock solution with a higher concentration.

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