The term "proof" is a spirits industry term that is defined as twice the % by volume of pure ethanol in solution. Thus, a solution that is 95% (by volume) ethanol is 190 proof. What is the molarity of ethanol in a vermouth drink, if the vermouth is 36 proof? Assume that the density of ethanol is 0.78 g/mL.

Relevance

the vermouth is 36 proof, so % by volume is 18% EtOH.

density = mass/ volume

0.78 g/mL = 0.78 / 0.001 L

780 g/L * 0.18 = mass

140.4 grams = mass

Ethanol is CH3CH2OH.

C: 2(12) = 24

H: 6 (1) = 6

O: 1(16) = 16

46 grams/mol

140.4 grams / 46 grams /mol = 3.05 mols

• Atomic weights: C=12 H=1 O=16 C2H6O=46

36 proof = 18% C2H6O (ethanol)

Let ethanol be called E. Let vermouth be called V.

18mLE/100mLV x 0.78gE/1mLE x 1molE/46gE x 1000mLV/1LV = 3.05 mole ethanol per 1L vermouth (solution) (molarity) to three significabnt figures

• Anonymous

If it's 36 proof then it has 18% ethanol. Assume you have a 100 ml (.100L)drink. 18 ml is ethanol and 82 ml is water. 18 ml x .78 g/ml = 14.04 g of ethanol.

Now if 1 mole of ethanol is 46 g, then you have to get the moles: 18g x 1 mol/46 g = 0.305 moles of ethanol

Molarity = Moles/ liter = .305/.100 = 3.05M

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• Given:

MW et. = 46 g / mol

Density = 0.78 g / ml.

% vol. = 18%

Find: molarity et. = moles et. / Liter vermouth solution

Assume 1 Liter vermouth solution = 1000 ml.

Volume et. = 18% x 1000ml = 180 ml.

Mass et. = 180 ml. x 0.78 g / ml = 140.4 g et.

moles et. = 140.4 g et. / 46 g / mol = 3.05 moles

Molarity et. = 3.05 moles / 1 Liter solution = 3.05