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if we did not use an excess of the BaCl2 solution it would decrease the mass percentage of sulfate in the unknown sample.
The net precipitation equation would be.
Ba2+(aq) + SO42-(aq) → BaSO4(s)
If BaCl2 (Ba2+) is not taken in excess then the precipitation would not be completed as some of the sulfate ions would still be remaining in the solution. This would decrease the mass percentage of sulfate in the unknown sample.
If some tiny pieces of filter paper still remained mixed with the precipitate(BaSO4) then the mass of sulfate would increase and it gives a high mass percentage of the sulfate.
mass percentage of sulfate = (mass of sulfate/mass of sample)*100
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Answer:
1 M
Explanation:
The molarity of a solution, M, is a measure of the concentration of that solution and it refers to the number of moles of solute (mol) per liter (L) of solution. The molarity (M) can be calculated using the formula:
M = number of moles (n) /volume (V)
In this question, a 500 ml aqueous solution of Na3PO4 was prepared using 82g of the solute.
Molar mass of Na3PO4 = 23(3) + 15 + 16(4)
= 69 + 31 + 64
= 164g/mol
Mole = mass/molar mass
mole = 82/164
mole = 0.5 mol
Volume in Litres (L) = 500 ml ÷ 1000 = 0.500L
Therefore, Molarity (M) = 0.5/0.500
Molarity = 1 M or 1 mol/L