The density of acetonitrile (CH3CN)is 0.786 g/mL, and the density of methanol (CH3OH)is 0.791 g/mL. A solution is made by dissolving 24.5mL (CH3OH)in 98.9mL (CH3CN). What is the mole fraction of methanol in the solution?
Use density to convert mL methanol and acetonitrile to grams.
Convert grams of each to mols.
mole fraction methanol= mols methanol/total mols.
mole fraction acetonitrile = mols acetonitrile/total mols.
To find the mole fraction of methanol in the solution, we need to determine the moles of methanol (CH3OH) and the moles of acetonitrile (CH3CN) in the mixture.
Step 1: Convert the volume of methanol (CH3OH) and acetonitrile (CH3CN) into mass using the given densities.
Mass of CH3OH = Volume of CH3OH × Density of CH3OH
= 24.5 mL × 0.791 g/mL
= 19.3455g
Mass of CH3CN = Volume of CH3CN × Density of CH3CN
= 98.9 mL × 0.786 g/mL
= 77.7054g
Step 2: Calculate the moles of CH3OH and CH3CN using their molar masses.
Molar mass of CH3OH = 12.01 g/mol (C) + 3(1.01 g/mol) (H) + 16.00 g/mol (O)
= 32.04 g/mol
Molar mass of CH3CN = 12.01 g/mol (C) + 3(1.01 g/mol) (H) + 14.01 g/mol (N)
= 41.05 g/mol
Moles of CH3OH = Mass of CH3OH / Molar mass of CH3OH
= 19.3455g / 32.04 g/mol
= 0.6034 mol
Moles of CH3CN = Mass of CH3CN / Molar mass of CH3CN
= 77.7054g / 41.05 g/mol
= 1.8926 mol
Step 3: Calculate the mole fraction of CH3OH in the solution.
Mole fraction of CH3OH = Moles of CH3OH / (Moles of CH3OH + Moles of CH3CN)
= 0.6034 mol / (0.6034 mol + 1.8926 mol)
= 0.2413
Therefore, the mole fraction of methanol (CH3OH) in the solution is 0.2413.