If 23 g of ethanol (c2h5oh) is dissolved in 60 g of ethenoic acid (ch3cooh). Then find out the mole fraction of each component?
1st step: figure how many moles of each you have.
Then figure the fraction.
To find the mole fraction of each component, we need to determine the number of moles of ethanol and ethanoic acid.
First, we will calculate the moles of ethanol (C2H5OH):
Molar mass of C2H5OH = 2 * molar mass of C + 6 * molar mass of H + 1 * molar mass of O
= (2 * 12.01 g/mol) + (6 * 1.008 g/mol) + (1 * 16.00 g/mol)
= 46.07 g/mol
Number of moles of ethanol = mass of ethanol / molar mass of ethanol
= 23 g / 46.07 g/mol
= 0.4997 mol (approximately)
Now, let's calculate the moles of ethanoic acid (CH3COOH):
Molar mass of CH3COOH = (1 * molar mass of C) + (2 * molar mass of H) + (2 * molar mass of O)
= (1 * 12.01 g/mol) + (2 * 1.008 g/mol) + (2 * 16.00 g/mol)
= 60.05 g/mol
Number of moles of ethanoic acid = mass of ethanoic acid / molar mass of ethanoic acid
= 60 g / 60.05 g/mol
= 0.9995 mol (approximately)
Now, we can calculate the mole fraction of each component:
Mole fraction of ethanol (Xethanol) = moles of ethanol / total moles in the solution
= 0.4997 mol / (0.4997 mol + 0.9995 mol)
= 0.333 (approximately)
Mole fraction of ethanoic acid (Xethanoic acid) = moles of ethanoic acid / total moles in the solution
= 0.9995 mol / (0.4997 mol + 0.9995 mol)
= 0.667 (approximately)
Therefore, the mole fraction of ethanol (C2H5OH) is 0.333, and the mole fraction of ethanoic acid (CH3COOH) is 0.667 in the given solution.