Answer is: molarity of calcium chloride is 0.123 M.
V₁(CaCl₂) = 165 mL; initial volume of calcium chloride solution.
c₁(CaCl₂) = 0.688 M; initial concentration.
V₂(CaCl₂) = 925.0 mL; final volume.
c₂(CaCl₂) = ?; final concentration.
Use c₁V₁ = c₂V₂.
0.688 M · 165 mL = c₂(CaCl₂) · 925 mL.
c₂(CaCl₂) = (0.688 M · 165 mL) ÷ 925 mL.
c₂(CaCl₂) = 0.123 M.
Answer:
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Answer: D) 2.37°C
Explanation:
