drbob you helped me with this problem the other day but i have a question.
-i changed 500mL of ethanol to grams then to mol. i was wondering if i then used 1000mL or 500mL of water to change to kg?
The "proof" of an alcoholic beverage is twice the volume percent of ethanol, C2H5OH, in water. The density of ethanol is 0.789g/mL and that of water is 1.00g/mL. A bottle of 100-proof rum is left outside on a cold winter day. a) Will the rum freeze if the temperature drops to -18 degrees C?
b) Rum is used in cooking nd baking. At what temperature does 100-proof rum boil?
I can't remember the details of the problem and my answer. I think the alcoholic beverage was 100 proof which means 50% v/v (volume/volume). That means we have 500 mL ethanol + 500 mL water (assuming water and alcohol volumes are additive and they are not), so you convert 500 mL ethanol to grams and that to mols. Then you convert 500 mL H2O to grams and that to kg. Then you can use delta T = kf*m. Does that answer your question. I'm sorry to have taken so long but I didn't remember what we did so I answered some other questions while I whirled this one around in my head. I hope I didn't give bad advice on both. :0)
No, your advice was great and this did answer my question. Thanks so much!
To convert the volume of ethanol to grams and then to moles, you can follow these steps:
1. Convert the volume of ethanol to grams:
- Given that the density of ethanol is 0.789 g/mL, you can use the following conversion: 500 mL ethanol x 0.789 g/mL = 394.5 grams of ethanol.
2. Convert the grams of ethanol to moles:
- You'll need to know the molar mass of ethanol, which is 46.07 g/mol.
- Use this equation: moles = grams/molar mass
- So, moles of ethanol = 394.5 g / 46.07 g/mol ≈ 8.56 moles of ethanol.
Now, you want to calculate the volume of water needed to convert the moles of ethanol to kilograms. To do this, you need to know the density of water and the molar mass of water.
The density of water is given as 1.00 g/mL. To convert this to kg/L, you can multiply by 1 g/mL * (1 kg/1000 g) * (1000 mL/1 L) = 1 kg/L.
The molar mass of water is 18.015 g/mol.
To calculate the volume of water needed, you can use the following equation: volume = (moles of ethanol x molar mass of water) / density of water.
Considering if you use 1000 mL (1 L) or 500 mL of water:
- If you use 1000 mL (1 L) of water, the volume of water needed will be (8.56 moles x 18.015 g/mol) / 1 kg/L ≈ 154.2 mL.
- If you use 500 mL of water, the volume of water needed will be (8.56 moles x 18.015 g/mol) / 1 kg/L ≈ 77.1 mL.
So, if you want to convert the moles of ethanol to kilograms and you started with 500 mL of ethanol, you would need approximately 77.1 mL of water.
Regarding your second question about the freezing and boiling points of rum, I can help with that as well. Please give me a moment to calculate the answers for you.