Respuesta :
0.04L is needed to precipitate 0.7 l of 0.13 m BaCl₂.
We utilize calculations to determine the number of moles for a particular molarity of solution.
Molarity is calculated as moles of solute per volume of solution (in litre)
Now, the molarity of the solution for the given barium ions is 0.13.
Solution volume is 0.7L.
When all the values are substituted, we get Moles of Barium = 0.7 0.13 = 0.091.
Chemical formula BaSO₄ » Ba²⁺⁺
According to the equation's stoichiometry, 1 mole of sulphate ions precipitates 1 mole of barium ions.
As a result, 0.091 moles of sulphate ions and 0.091 moles of barium ions precipitate.
Sulfate ion volume calculation:
Now, use the aforementioned equation to determine the volume of ions.
Sulfate ion per mole:
0.091 moles
Sulfate ion molarity concentration = 0.05 m
By substituting all value ,
We get
C = n/V
V = n/ C= 0.091 / 0.5 = 0.04L
As a result, we determined that 0.04L is needed to precipitate 0.7 l of 0.13 m BaCl₂.
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