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leva [86]
2 years ago
5

What volume would be needed to prepare 375 mL of a .45 M CaCl2 using only a solution of 1.0 M CaCl2 and water?

Chemistry
1 answer:
Neporo4naja [7]2 years ago
8 0

Answer:

168.75 ml

Explanation:

M1V1=M2V2

375ml*.45M=1M*V2

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