A = grams acetone
E = grams ethanol
dA = density acetone
dE = density ethanol
VA = volume acetone
VE = volume ethanol
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You want the volume to be 1000 mL; therefore, VA + VE = 1000 mL but since volume = mass/density, then
A/dA + E/dE = 1000....eqn 1
Two unknowns means you need two equations. Equation 2 you want mols to be equal. mols = grams/molar mass so
A/58 = E/46.....eqn 2.
Solve for A and E which gives you grams A and grams E.
I scratched through the problem quickly and the values are close, but just estimates of about 350 g for ethanol and 440 g for acetone but you need to confirm those since I think these are just close answers.
calculate the volumes of acetone and ethanol that was used to mix together in order to produce 1 litre (1000ml) of equimolar mixture. helpful notes are, ethanol Mr=46 and a density (20 degrees C)=0.789g/ml, acetone Mr=58 and density (20 degrees C)=0.793g/ml
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