How much aluminum chloride is required to prepare a 0.2 M solution with a volume of 0.5 liters?

Determine the required amount of substance to prepare the solution.
To do this, multiply the volume of the solution by its molar concentration.
In order to prepare such a solution, it is necessary to weigh and dissolve 0.2 x 0.5 = 0.1 mol of salt in water.
Let’s calculate the required weight of salt.
To do this, multiply the molar mass of the salt by the chemical amount of the substance.
M AlCl3 = 27 + 35.5 x 3 = 133.5 grams / mol;
m AlCl3 = 0.1 x 133.5 = 13.35 grams;

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