Question

Three containers of hydrates used to this experiment were mislabeled. We know that the containers were labelled originally NaHSO4•H2O, Na2SO4•2H2O and Na2SO4•10H2O. How would you determine the actual contents of each container?

Copper(II) sulfate can form several hydrates: CuSO4•H2O, CuSO4•3H2O and CuSO4•5H2O. You got a sample of an unknown copper hydrate. You remove the water by heating the hydrate to a constant mass. The mass percent of removed water was 25.3 %. Which hydrate was it?

Answer 1

Solo Given: There are three containers of Hydrates were labelled originally NaHS04. H20 ; Na₂504 . 2420 ; Na₂504. 10420 If we have determine the actual contents of each container. We can determine it by heating it to remove water of crystallization Nanson. H20 Narlson Na, 504.2420 A Na2504 Na₂504. 101/20 → Na₂504 Ala molar mass of Nansoy. 120 (23+1+96 +18) =138 qmimol to content of water = 18 X 100) 1 = 13.04 % 138 » molar mass of Na2504. 2420 (46+96+36) - 178 gmlmol 1. content of water = 36 8100 = 20.224. 178 a molar mass of Naz 5o4. 10420 dag stansen = 46+ 96+ 180 = 322 gm /mol 1. content of water a 1180 x 100 y = 55.90%. (322

By calculating determine the the removed water / we can actual content of the container 27. Given: we got - mass t. of sample of unknown removed water copper hydrate = 25.3.1. Sample 1 - Lusoa. H2O cusot molar mass of cuson. 10 2 (64 +96718) 2 178 (mimoll f water content = 18 X 100 178 210.4.1. sample 2 Cuso4. 3H20 A cusor molar mass of custa. 3420 =(64+96 +54) = 214 gmimo) of water content = 54 X 100 214 Bd. 25.31. Cu S04.5120 Sample 3 Cuso4 molwe mass 06 CuS04, 5420 2 (64+ 96 +90) = 250 gm/mol 't water content 2 90 x 100 = 37. 51. 250 Hence; required sample was cuso4, 3H20 which removes 25.3% of wata of Hallisation
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