A saturated solution of NaNO3 is prepared at 60.°C using 100. grams of water. As this solution is cooled to 10.°C, NaNO3 precipitates (settles)
out of the solution. The resulting solution is saturated. Approximately how many grams of NaNO3 settled out of the original solution?
(1) 46 g (3) 85 g
(2) 61 g (4) 126 g



Answer :

The answer is (1) 46g. This is related to the solubility curve of the NaNO3. The solubility of NaNO3 under 60 ℃ is 126 g. And the solubility under 10 ℃ is 80 g. So the NaNO3 precipitates is 126-80=46 g.

The answer is 46 g


The explanation :


According to the attached curve of the relation between the temperature and the grams in solutions:


-we can see from the curve that the grams of NaNO3 at 60°C = 126 g


-and the grams of NaNO3 at 10°C = 80 g


-so, To get the grams of NaNO3 settled out of the original solution, we will subtract the grams of NaNO3 at 10°C - the grams of NaNO3 at 60°C


126 g - 80 g = 46 g


∴ the correct answer is 46 g

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