Assuming that the sodium sulphate sample is 100% pure, calculate the volume of 0.25 M BaCL2 required to precipitate all the sulphate in it. (as in step 4 in the precipitation stage in the experiment). What is the formula use for this question ? thank you.
1 mol X mol
Na2SO4 +BaCl2 = BaSO4↓ + 2NaCl
1 mol 1 mol
n(Na2SO4) = 1 mol = 142 g/mol.
n(BaCl2) = X = 1 mol.
V(BaCl2) = n(BaCl2) / C(BaCl2) = 1 mol / 0.25 M = 4 L.
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