Answer to Question #323876 in General Chemistry for Rose

Question

2. A solution us prepared by mixing 1g of ethanol ( C2H6O) with 100g water to give a final volume of 101 ml. Calculate the mole fraction for the solute and solvent,and the molarity of ethanol in the solution.

Answer 1

n(C₂H₆O) = m(C₂H₆O)/Mr(C₂H₆O) = 1 g / 46.07 g/mol = 0.0217 mol

n(H₂O) = m(H₂O)/Mr(H₂O) = 100 g / 18.015 g/mol = 5.5509 mol

n(total) = 0.0217 mol + 5.5509 mol = 5.5726 mol

$\chi$_ethanol = n(C₂H₆O) / n(total) = 0.0217 mol / 5.5726 mol = 0.0039

$\chi$_water = n(H₂O) / n(total) = 5.5509 mol / 5.5726 mol = 0.9961

Molarity = Moles of solute / Liters of solution

Molarity = n(C₂H₆O) / V(solution) = 0.0217 mol / 0.101 L = 0.215 M