Question

Solution A contained 40 cm3 volume of wine diluted to 100 cm3 using deionized water. Solution B contained 1 cm3 taken from Solution A and diluted to 10 cm3 with deionised water.  After running Solution B through the GC it was found it contained 0.045 cm3 of ethanol. What is the % alcohol by volume (ABV) in the original wine sample?

Answer 1

Ans. Let total amount of ethanol in 40.0 mL of wine be X cm³.

# [Ethanol] in solution A:

Using C1V1 (wine) = C2V2 (solution A)                ; where, C = volume of ethanol

            Or, X cm³ x 40.0 cm³ = C2 x 100.0 cm³

            Or, C2 = (X cm³ x 40.0 cm³) / 100.0 cm³

            Hence, C2 = 0.4X cm³

Therefore, net content of ethanol in 100.0 cm³ solution A = 0.4X cm³

# [Ethanol] in solution B:

Using C1V1 (solution A) = C2V2 (solution B)                  ; where, C = volume of ethanol

            Or, 0.4X cm³ x 1.0 cm³ = C2 x 10.0 cm³

            Or, C2 = (0.4X cm³ x 1.0 cm³) / 10.0 cm³

            Hence, C2 = 0.04X cm³

Therefore, net content of ethanol in 10.0 cm³ solution B = 0.04X cm³

# From GC, the net content of ethanol in solution B is 0.045 cm³.

So,

            0.04X cm³ = 0.045 cm³

            Or, X = 0.045 / 0.04

            Hence, X = 1.125

Therefore, total amount of ethanol in 40.0 mL of wine be 1.125 cm³.

# So far, we have,

            Volume of ethanol in original wine solution = 1.125 cm³

            Total volume of wine = 40.0 cm³

Now,

            % (v/v) ethanol in wine = (Volume of ethanol / Volume of Wine) x 100

                                                = (1.125 cm³ / 40.0 cm³) x 100

                                                = 2.81 %