The mass fraction of impurities in limestone is 6.25%. Calculate the mass of the basic substance (calcium carbonate)
The mass fraction of impurities in limestone is 6.25%. Calculate the mass of the basic substance (calcium carbonate) contained in 1.5 tons of natural limestone.
Given: ω (impurities) = 6.25%, m (sample) = 1.5 tons.
Find: ω (basic substance) -?
Decision.
1. Let’s calculate the mass of impurities in a sample of natural limestone:
m (impurities) = (ω (impurities) • m (sample)): 100% = 6.25% • 1.5 t: 100% = 0.09375 t.
2. Calculate the mass of the main substance:
m (basic substance) = m (sample) – m (impurities) = 1.5 t – 0.09375 g = 1.40625 t.
2. Let’s calculate the mass fraction of the main substance:
ω (basic substance) = m (basic substance) / m (sample) = 1.40625: 1.5 = 0.9375, or,
multiplying this number by 100%, we get 93.75%.
Answer: 93.75%.
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