(chemistry)
23g of ethanol was mixed with 72g of water in a reaction vessel.calculate the mole fraction and mole percent of water.calculate the mole fraction and mole percent of ethanol
To calculate the mole fraction, we need to determine the moles of each component in the mixture.
Step 1: Calculate the moles of ethanol (C2H5OH):
We know the molar mass of ethanol (C2H5OH) is 46.07 g/mol.
Moles of ethanol = mass of ethanol / molar mass of ethanol
= 23 g / 46.07 g/mol
Step 2: Calculate the moles of water (H2O):
We know the molar mass of water (H2O) is 18.02 g/mol.
Moles of water = mass of water / molar mass of water
= 72 g / 18.02 g/mol
Step 3: Calculate the mole fraction of each component:
Mole fraction of ethanol = moles of ethanol / total moles
= moles of ethanol / (moles of ethanol + moles of water)
Mole fraction of water = moles of water / total moles
= moles of water / (moles of ethanol + moles of water)
Step 4: Calculate the mole percent:
Mole percent = mole fraction * 100
Now, let's calculate the mole fraction and mole percent:
Moles of ethanol = 23 g / 46.07 g/mol = 0.4991 mol (approximately)
Moles of water = 72 g / 18.02 g/mol = 3.9989 mol (approximately)
Mole fraction of ethanol = 0.4991 mol / (0.4991 mol + 3.9989 mol) = 0.111
Mole fraction of water = 3.9989 mol / (0.4991 mol + 3.9989 mol) = 0.889
Mole percent of ethanol = 0.111 * 100 = 11.1%
Mole percent of water = 0.889 * 100 = 88.9%
Therefore, the mole fraction of ethanol is 0.111 or 11.1%, and the mole fraction of water is 0.889 or 88.9%.