
В Воде объемом 3л растворен аммиак объемом 56л (н.у). Вычислите массовую долю аммиака в растворе.


Ответы на вопрос

Масса аммиака = V/Vm*M = 56/22.4*17 = 42,5 г
Масса раствора = 3000+42,5 = 3042,5 г
ω(аммиака) = 42,5/3042,5*100 = 1,397%



Calculation of the Mass Fraction of Ammonia in the Solution
To calculate the mass fraction of ammonia in the solution, we need to know the volume of water and the volume of ammonia in the solution.
Given: - Volume of water = 3 liters - Volume of ammonia = 56 liters (at standard temperature and pressure)
To calculate the mass fraction, we can use the formula:
Mass Fraction = (Mass of Ammonia) / (Mass of Solution)
To find the mass of ammonia, we need to know the density of ammonia. Unfortunately, the search results did not provide the necessary information to calculate the mass fraction accurately.
However, if we assume that the density of ammonia is approximately 0.681 kg/L at standard temperature and pressure, we can estimate the mass of ammonia in the solution.
Using this assumption, we can calculate the mass of ammonia as follows:
Mass of Ammonia = Volume of Ammonia * Density of Ammonia
Mass of Ammonia = 56 L * 0.681 kg/L
Mass of Ammonia = 38.136 kg
Now, we can calculate the mass fraction of ammonia:
Mass Fraction = Mass of Ammonia / (Mass of Ammonia + Mass of Water)
Mass Fraction = 38.136 kg / (38.136 kg + Mass of Water)
Since the volume of water is given as 3 liters, we can assume the density of water to be approximately 1 kg/L. Therefore, the mass of water is:
Mass of Water = Volume of Water * Density of Water
Mass of Water = 3 L * 1 kg/L
Mass of Water = 3 kg
Substituting the values into the mass fraction formula:
Mass Fraction = 38.136 kg / (38.136 kg + 3 kg)
Mass Fraction = 38.136 kg / 41.136 kg
Mass Fraction ≈ 0.926 or 92.6%
Therefore, the estimated mass fraction of ammonia in the solution is approximately 92.6%.
Please note that this calculation assumes the density of ammonia and water at standard temperature and pressure. The actual density values may vary depending on the specific conditions.


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