Question

The alcohol content of hard liquor is normally given in terms of the "proof," which is defined as twice the percentage by volume of ethanol (C2H5OH) present. Calculate the number of grams of alcohol present in 1.00 L of 48-proof gin. The density of ethanol is 0.798 g/mL.

Answer 1

m C2H5OH = 191.52 g

Step-by-step explanation:

• proof ≡ (2)(%v/v C2H5OH)

∴ %v/v C2H5OH = ( v C2H5OH / v sln)×100

⇒ 48 proof = (2) (%v/v C2H5OH)

⇒ 48/2 = %v/v C2H5OH

⇒ 24 = %v/v C2H5OH

⇒ 24/100 = v C2H5OH / v sln = 0.24

∴ v sln (gin) = 1.00 L

⇒ v C2H5OH = ( 0.24 )( 1.00 L )

⇒ v C2H5OH = 0.24 L = 240 mL

⇒ m C2H5OH = (240 mL)(0.798 g/mL)

⇒ m C2H5OH = 191.52 g