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weqwewe [10]
3 years ago
15

An experiment requires 15.0 ml of a 0.100 m solution of hcl. if you have a 12.0 m hcl stock solution in the laboratory, how woul

d you prepare enough 0.1 m solution to run at least 3 experiments?
Chemistry
1 answer:
soldi70 [24.7K]3 years ago
6 0
Answer is: A. add 0.833 mL 12 M HCl stock solution to 99.167 mL water. 

Missing question: 

A. add 0.833 mL 12 M HCl stock solution to 99.167 mL water. 

B. add 0.833 mL 12 M HCl stock solution to 49.167 mL water. 

C. add 0.833 mL 12 M HCl stock solution to 199.167 mL water.


c₁ - original concentration of the solution, before it gets diluted.

c₂<span> - final concentration of the solution, after dilution.
V</span>₁<span> - volume to be diluted.
V</span>₂<span> - final volume after dilution.
c</span>₁ · V₁ = c₂ · V₂<span>.
</span>c₁(HCl) = 12.0 M.

c₂(HCl) = 0.100 M.

V₂(HCl) = 3 · 15.0 mL = 45.0 mL; for three experiments.

V₁(HCl) = c₂ · V₂ / c₁.

V₁(HCl) = 0.375 mL.

Make proportion: 0.375 mL : 0.833 mL = 45 mL : V.

V = 100 mL.


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