calculate the number of grams of alcohol present in 1.70 L OF 75-Proof gin. The density of ethanol is 0.798 g/ml. Proof--> twice the % by volume of ethanol(C2H5OH) present.
75 proof = 37.5% v/v
That means 37.5 mL ethanol/100 mL sample. You have 1700 mL; therefore, you have
37.5 x (1700/100) = ?? mL ethanol.
Use the density to convert to grams.
To calculate the number of grams of alcohol present in 1.70 L of 75-Proof gin, you need to follow a few steps:
Step 1: Understand the concept of "proof" and how it relates to the percentage of alcohol by volume (ABV):
- Proof is a measure of the alcohol content in a beverage and is usually twice the percentage of alcohol by volume (ABV).
- For example, if a gin is 50-proof, it means it contains 25% ABV (50 divided by 2).
Step 2: Calculate the ABV of the gin:
- Since the gin is 75-proof, you can divide the proof value by 2 to get the ABV.
- 75-proof gin = 75/2 = 37.5% ABV.
Step 3: Use the ABV and the provided volume to calculate the number of milliliters (ml) of alcohol:
- Multiply the volume in liters by 1000 to convert it to milliliters (since 1 liter = 1000 ml).
- 1.70 L x 1000 ml/L = 1700 ml.
- Multiply the volume in milliliters by the ABV to get the amount of alcohol in milliliters.
- 1700 ml x 37.5% = 637.5 ml of alcohol.
Step 4: Convert the milliliters of alcohol to grams:
- Multiply the volume in milliliters by the density of ethanol to get the mass in grams.
- 637.5 ml x 0.798 g/ml = 508.95 grams of alcohol.
Therefore, there are approximately 508.95 grams of alcohol present in 1.70 liters of 75-Proof gin.