1.285g sample of Na2SO4 is exposed to the atmosphere and found to gain 0.393g in

Question

1.285g sample of Na2SO4 is exposed to the atmosphere and found to gain 0.393g in mass. What is the percent, by mass, of Na2SO4*H2O in the resulting mixture of anhydrous Na2SO4 and the decahydrate?

Explanation / Answer

i'm not sure of what you are asking since you want the % of Na2SO4-10H2O in the mixture of Na2SO4-?H2O. i'm going to solve this as i would have asked it. 1.263g Na2SO4 + 0.383g H2O -->1.646g Na2SO4 - 10H2O in the known compound of Na2SO4-10H2O, the % of water in 1 g is the same as the % of water in 2 grams and the same % of water in 1 mole g10H2O / gNa2SO4-10H2O x 100% = ? 180gH2O / 322g Na2SO4-10H2O x100% = 55.9% water from the data, 0.383gH2O / 1.646gNa2SO4-xH2O = 23.3% therefore, the compound that is made by the absorption of water in your experiment is not equal to the % of water in Na2SO4-10H2O

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