Please help e solve this and show me how you did it

Question

Students attempted to distill ethanol from liquors and spirits to help pandemic frontliners in the problem of disinfectant-shortage . Three-time redistillation was conducted to generate 20 L distillate with a density of 0.814 g/mL at 25 °C. Using the table below, calculate the amount of ethanol in the recovered distillate. Density of pure ethanol is 0.789 g/mL. 

Answer 1

Well from the density given and the table 90% of the mass of the ethanol-water mixture must be ethanol.

So in the 20L lets assume we have x ml of ethanol. Then we have 20000-x ml of water.

Mass of ethanol = 0.789x

Mass of water = (20000-x) x 0.001 (1 ml of water has mass 1g = 0.001 kg)

Total mass = 0.789x + (20000-x) x 0.001

Ethanol is 90% so [0.789x/[0.789x + 0.001(20000-x)] = 0.90

You can solve this for x, then knowing x you can calculate the mass of ethanol.

0.789x = 0.7101x + 10 - 0.0009x

0.789x = 0.711x + 10

0.078x = 10

x = 128.2 ml

mass = 128.2 x 0.789 = 101.2 g