Question

What is the molarity of a solution that was prepared by dissolving 82 g of cacl2?

Answer 1

The molarity of a solution is 0.909 M.

m(CaCl₂) = 82 g; mass of calcium chloride

M(CaCl₂) = 111 g/mol; molar mass of calcium chloride

n(CaCl₂) = m(CaCl₂) ÷ M(CaCl₂)

n(CaCl₂) = 82 g ÷ 111 g/mol

n(CaCl₂) = 0.74 mol; amount of calcium chloride

V(CaCl₂) = 812 mL = 0.812 L; volume of calcium chloride

c(CaCl₂) = n(CaCl₂) ÷ V(CaCl₂)

c(CaCl₂) = 0.74 mol ÷ 0.812 L

c(CaCl₂) = 0.909 mol/L = 0.909 M; molarity of calcium chloride

Calcium chloride is a inorganic salt with ionic bonds between calcium and chlorine. It is a colorless crystalline solid at room temperature, highly soluble in water.

More about molarity: brainly.com/question/26873446

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