200 g of water was added to 200 ml of a 96% ethanol solution (density 0.8 g / ml). Determine the mass fraction (percentage) of alcohol in the new solution.
1. Let’s find the mass of the С2Н5ОН solution:
m (solution of С2Н5ОН) = V * ρm = 200 * 0.8 = 160g.
2. Let’s find the mass of pure С2Н5ОН:
m (C2H5OH) = 96 * 160/100 = 153.6 g.
3. Let’s find the mass fraction of С2Н5ОН:
ω (C2H5OH) = 153.6 * 100 / (160 + 200) = 42% (0.42).
Answer: ω (C2H5OH) = 42% (0.42).
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