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
To calculate the volumes of acetone and ethanol needed to produce a 1-liter (1000 ml) equimolar mixture, we can use the following steps:
Step 1: Determine the molar amounts of acetone and ethanol required.
Since we want to create an equimolar mixture, which means both compounds have an equal molar concentration, we need to find the moles of each substance.
To find the moles, we can use the formula:
moles = mass / molar mass
For ethanol (C2H5OH):
Molar mass of ethanol (C2H5OH) = 46 g/mol (given)
We don't know the mass of ethanol yet, so we can calculate it using the density and volume:
mass of ethanol = density × volume
Given density of ethanol at 20 degrees C: 0.789 g/ml
Volume of ethanol = 1000 ml (1 liter)
Calculating mass of ethanol:
mass of ethanol = 0.789 g/ml × 1000 ml = 789 g
Now we can calculate the moles of ethanol:
moles of ethanol = mass of ethanol / molar mass of ethanol
moles of ethanol = 789 g / 46 g/mol
For acetone (CH3COCH3):
Molar mass of acetone (CH3COCH3) = 58 g/mol (given)
Using the same process, we can calculate the moles of acetone. First, we need to find the mass of acetone:
mass of acetone = density × volume
density of acetone at 20 degrees C: 0.793 g/ml
volume = 1000 ml (1 liter)
Calculating mass of acetone:
mass of acetone = 0.793 g/ml × 1000 ml = 793 g
Now we can calculate the moles of acetone:
moles of acetone = mass of acetone / molar mass of acetone
moles of acetone = 793 g / 58 g/mol
Step 2: Determine the volumes of acetone and ethanol based on their calculated moles.
Since we have the moles of each substance, we can use the molar volume to calculate their individual volumes.
The molar volume is defined as the volume occupied by one mole of any substance. For gases, it is known to be approximately 22.4 liters/mole under standard conditions. However, since acetone and ethanol are liquids, we need to use their densities to calculate their molar volumes.
Molar volume of ethanol (C2H5OH) = 46 g/mol / 0.789 g/ml = 58.4 ml/mol
Molar volume of acetone (CH3COCH3) = 58 g/mol / 0.793 g/ml = 73.3 ml/mol
Now, we can calculate the volumes of acetone and ethanol needed for a 1-liter (1000 ml) equimolar mixture.
Volume of ethanol = moles of ethanol × molar volume of ethanol
Volume of acetone = moles of acetone × molar volume of acetone
Volume of ethanol = (789 g / 46 g/mol) × 58.4 ml/mol
Volume of acetone = (793 g / 58 g/mol) × 73.3 ml/mol
Calculating these values will give you the specific volumes of ethanol and acetone required to produce a 1-liter equimolar mixture.
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.