How would you prepare 250. mL of a 0.175 M solution of Cobalt(II) Chloride from the solid and distilled water?

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How would you prepare 250. mL of a 0.175 M solution of Cobalt(II) Chloride from the solid and distilled water?

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Answer 1 · 2018-03-15T00:56:27

Consider that solutions are solutes dissolved in solvent.

Here, our solute is cobalt (II) chloride and our solvent is distilled water.

The volume of solute is generally ignored, so we would dissolve,

$250 mL \cdot \frac{L}{10^{3} mL} \cdot 0.175 M \cdot \frac{129.9 g}{mol} \approx 5.68 g$ $250"mL" * "L"/(10^3"mL") * 0.175"M" * (129.9"g")/"mol" approx 5.68"g"$

of cobalt (II) chloride in approximately $250 mL$ $250"mL"$ of distilled water to create the solution in the lab.

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How would you prepare 250. mL of a 0.175 M solution of Cobalt(II) Chloride from the solid and distilled water?

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