A 0.5 m^3 container is filled with a mixture of 10% by volume ethanol and 90% by volume water at 25 °C. Find the weight of the liquid.

total weight of liquid = 4788.25 N or 488.09 kg

Explanation:

given data

total volume = 0.5 m³

volume of ethanol = 10 % of volume = 0.10 * 0.5 = 0.05 m³

volume of water = 90 % at 25 °C of volume = 0.90 * 0.5 = 0.45 m³

to find out

weight of the liquid

solution

we know that density of water at 25 is 997 kg/m³

and density of ethanol is 789 kg/m³

so weight of water is = density * volume * g

put here value and we take g = 9.81

weight of water is = 997 * 0.45 * 9.81

weight of water = 4401.25 N ... 1

weight of ethanol is = density * volume * g

put here value and we take g = 9.81

weight of ethanol is = 789 * 0.05 * 9.81

weight of ethanol = 387.00 N ... 2

so total weight of liquid = sum of equation 1 add 2

total weight of liquid = 4401.25 + 387

total weight of liquid = 4788.25 N or 488.09 kg