Respuesta :
The volume required to precipitate 0.7l of 0.13 m BaCl2 is 0.04L.
Calculations of number of moles for given molarity of solution
we use
Molarity =Moles of solute / volume of solution (in litre)
Now, for barium ions
Given,
Molarity of solution = 0.13 m
Volume of solution = 0.7L
By substituting all the value ,we have
Moles of Barium = 0.7 × 0.13
= 0.091
Chemical equation
BaSO4 » Ba2+ + So2-
By stoichiometry of equation
1 mole of sulphate ion precipitate 1 mole of barium ion
So, 0.091 mole of sulphate ion precipitate
0.091 of moles of Barium ion
Calculation of volume of sulfate ion
Now, calculate the volume of ions by using equation as given above
Mole of sulphate ion = 0.091 mol
Molarity concentration of sulphate ions = 0.05m
By substituting all value , we get
C = n / V
V = n/ C
= 0.091 / 0.5 = 0.04L
Thus, we calculated that the volume required to precipitate 0.7l of 0.13 m BaCl2 is 0.04L.
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