The Preparation of 0.335 m LiCl Solution, how many grams of solid LiCl are needed to prepare 225 ml of 0.335 m solution

The Preparation of 0.335 m LiCl Solution

The preparation of a 0.335 m LiCl solution requires the calculation of the mass of solid LiCl needed to prepare the solution. LiCl, or lithium chloride, is an ionic compound composed of one atom of lithium and one atom of chlorine. It is a hygroscopic salt, meaning it absorbs moisture from the air and is typically found in crystalline form (Chaplin, 2019). To prepare a 0.335 m LiCl solution, a measured amount of LiCl must be weighed and dissolved in a given volume of water. The mass of LiCl required to prepare a 0.335 m solution can be calculated using the formula: mass (g) = molarity (mol/L) x volume (L) x molar mass (g/mol). In this case, the desired molarity is 0.335 mol/L, the volume is 0.225 L, and the molar mass of LiCl is 42.39 g/mol (Khan, 2020). Plugging these numbers into the formula results in a required mass of 3.2 g of LiCl. Thus, to prepare a 0.335 m LiCl solution with a volume of 0.225 L, 3.2 g of solid LiCl must be weighed and dissolved in the solution. Cont…

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