A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL. Calculate the concentration of ethanol in wine in terms of molarity and mass percent.

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 = > [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= { (2.2mol) (46g/mol) ]/1000g soln] x 100% = 10.1% (w/w) in EtOH.