The Preparation of 0.335 m LiCl Solution

The Preparation of 0.335 m LiCl Answer

The preparation of a 0.335 m LiCl answer requires the calculation of the mass of stable LiCl wanted to organize the answer. LiCl, or lithium chloride, is an ionic compound composed of 1 atom of lithium and one atom of chlorine. It’s a hygroscopic salt, which means it absorbs moisture from the air and is usually present in crystalline kind (Chaplin, 2019). To organize a 0.335 m LiCl answer, a measured quantity of LiCl should be weighed and dissolved in a given quantity of water. The mass of LiCl required to organize a 0.335 m answer will be calculated utilizing the formulation: mass (g) = molarity (mol/L) x quantity (L) x molar mass (g/mol). On this case, the specified molarity is 0.335 mol/L, the quantity is 0.225 L, and the molar mass of LiCl is 42.39 g/mol (Khan, 2020). Plugging these numbers into the formulation ends in a required mass of three.2 g of LiCl. Thus, to organize a 0.335 m LiCl answer with a quantity of 0.225 L, 3.2 g of stable LiCl should be weighed and dissolved within the answer.