A chemist makes 660.mL of iron(II) bromide FeBr2 working solution by adding distilled water to 90.mL of a 1.28molL stock solution of iron(II) bromide in water.
Calculate the concentration of the chemist's working solution. Round your answer to 2 sig figs.
Calculate the concentration of the chemist's working solution.?
2016-10-12 6:02 pm
回答 (3)
2016-10-12 6:24 pm
Before dilution (stock solution) : M₁ = 1.28 mol/L, V₁ = 90 mL
After dilution (working solution) : M₂ = ? mol/L, V₂ = 660 mL
No. of moles of FeBr₂ in stock solution = No. of moles of FeBr₂ in working solution
M₁V₁ = M₂V₂
(1.28 mol/L) × (90 mL) = M₂ × (660 mL)
Concentration of working solution, M₂ = 1.28 × 90 / 660 mol/L = 0.17 mol/L (to 2 sig. figs.)
After dilution (working solution) : M₂ = ? mol/L, V₂ = 660 mL
No. of moles of FeBr₂ in stock solution = No. of moles of FeBr₂ in working solution
M₁V₁ = M₂V₂
(1.28 mol/L) × (90 mL) = M₂ × (660 mL)
Concentration of working solution, M₂ = 1.28 × 90 / 660 mol/L = 0.17 mol/L (to 2 sig. figs.)
2016-10-12 6:22 pm
C1V1 = C2V2
1.28 mol/L X 90 mL = C1 X 550 mL
C1 = 0.17 M
1.28 mol/L X 90 mL = C1 X 550 mL
C1 = 0.17 M
2016-10-12 6:20 pm
(90.mL) x (1.28 mol/L) / (660.mL) = 0.17 mol/L
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