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Topic: 2018 VERIFIED CHEMISTRY PRACTICAL ANSWER NOW AVAILABLE

Post By:
On September 14, 2018, at: 4:12 am
Category: Education
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(1ai)
Tabulate:
Burette reading |1 |2 |3 |
final reading |24.80|34.50|20.65|
initial reading |0.00 |10.50|6.55 |
vol of acid used|24.80|24.00|24.10|
(1aii) Average volume of A used
Va = 24.00 + 24.10/2
= 24.05cm^3

(1bi)
Concentration of A in moldm^-3
A contains 0.79g of KMnO4 per 250cm³ of solution.
Hence since 250cm³ = 0.79g
100cm³ = Xg
X = 1000×0.79/250 = 3.16g
Hence Conc in g/dm³ of A = 3.16gdm^-³
But molar conc = mass conc(gdm^-³)/molar mass(g/mol)
Hence molar Conc.Of A= 3.16gdm^-3/molar mass ofA
But molar mass of A = KMnO4
= 39 + 55 + 4(16) = 39 + 55 + 64
= 158g/mol
Hence Conc in moldm^-³ of A = 3.16gdm^-3/158g/mol
= 0.020moldm^-3
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Category: Education

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