F4 mole 問題

[匿名]

2008-04-30 5:26 am

4.15g of a sample of hydrated sodium carbonate, of formula Na2CO3‧nH20, is dissolved in water and the solution made up to 25.0 cm3 . With methyl orange as indicator. 25.0cm3 of this solution requires 29.0cm3 of 0.05M sulphuric acid for neutralization. Calculate n, the number of molecules of water of crystallization in a formula unit of the sodium carbonate hydrate sample.

回答 (1)

Uncle Michael

2008-04-30 7:51 am

✔ 最佳答案

There is a typing mistake:
The solution is made up to 250 cm3, but not 25.0 cm3.

Consider the titration:
Na2CO3 + H2SO4 → Na2SO4 + H2O
Mole ratio Na2SO4 = 1 : 1
No. of moles of H2SO4 used = MV = 0.05 x (29/1000) = 0.00145 mol
No. of moles of Na2CO3 used = 0.00145 mol
No. of moles of Na2CO3•nH2O used = 0.00145 mol

25 cm3 of solution is used in titration out of 250 cm3.
Mass of Na2CO3•nH2O used in titration = 4.15 x (25/250) = 0.415 g
Molar mass of Na2CO3•nH2O = mass/mol = 0.415/0.00145 = 286 g mol-1

Molar mass of Na2CO3 = 23x2 + 12 + 16x3 = 106 g mol-1
Molar mass of H2O = 2x1 + 16 = 18 g mol-1
Molar mass of Na2CO3•nH2O = (106 + 18n) g mol-1 = 286 g mol-1
Hence, n = 10

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