What volume of 0.5M CaCl₂ solution is needed to prepare 250 ml of 0.5 ml of solution that has Cl^- concentration 0.1 M?

A 12.5 ml

B. 25 ml

C.50 ml

D.200 ml

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Solution

First the Molarity of the CaCl₂ from the molarity of the Cl⁻ ions is calculated.

Molarity of CaCl₂ in the concentration of 0.1M of Cl⁻ ionis is :-

0.1M / 2 = 0.05M of CaCl₂

since of M₁V₁ = M₂V₂

(Where M₁ is the molarity of the first solution,

V₁ is the volume of the first solution,

M₂ is the concentration of the second solution,

V₂ is the volume of the second solution)

The volume required is M₁V₁ = M₂V₂

0.5M × V₁ = 0.05M × 250/1000

V₁ = 0.05 × 0.25 / 0.5

= 0.025 L or 25 ml

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