Answer to Question #308882 in Chemistry for Junard

Question

Mole Fraction 1. Calculate the mole fraction of each component of a solution containing 65 g of ethanol (C₂H6O) in 350g of water.

Answer 1

Solution:

solute = ethanol (C₂H₆O)

solvent = water (H₂O)

The molar mass of ethanol (C₂H₆O) is 46.07 g/mol

The molar mass of water (H₂O) is 18.015 g/mol

Calculate the moles of each component:

Moles of C₂H₆O = (65 g C₂H₆O) × (1 mol C₂H₆O / 46.07 g C₂H₆O) = 1.411 mol C₂H₆O

Moles of H₂O = (350 g H₂O) × (1 mol H₂O / 18.015 g H₂O) = 19.428 mol H₂O

Therefore,

Total moles of solution = Moles of C₂H₆O + Moles of H₂O = 1.411 mol + 19.428 mol = 20.839 mol

Calculate the mole fraction (χ) of each component:

χ = Moles of component / Total moles of solution

Therefore,

χ_solute = χ_C2H6O = Moles of C₂H₆O / Total moles of solution = (1.411 mol) / (20.839 mol) = 0.0677

χ_C2H6O = 0.0677

χ_solvent = χ_H2O = Moles of H₂O / Total moles of solution = (19.428 mol) / (20.839 mol) = 0.9323

χ_H2O = 0.9323

Answer:

The mole fraction of ethanol is 0.0677

The mole fraction of water is 0.9323