200 g of water was added to 200 ml of a 96% ethanol solution (density 0.8 g / ml).

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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