The coefficient of thermal expansion of ethanol equals 1.12 × 10^(-3) K^(-1) at 20°C and 1.000 atm. The density at 20°C is equal to 0.7893 g/cm^(-3). Find the volume of 1.000 mol of ethanol at 10.00°C and 1.000 atm. The molecular weight of ethanol is 46.07 g/mol.
Added by Montserrat P.
Step 1
First, we need to find the initial volume of 1.000 mol of ethanol at 20°C and 1.000 atm. We can use the density and molecular weight to find this: Density = mass/volume 0.7893 g/cm³ = (1.000 mol * 46.07 g/mol) / volume volume = (1.000 * 46.07) / 0.7893 volume ≈ Show more…
Show all steps
Your feedback will help us improve your experience
Supreeta N and 73 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Calculate the pressure of ethanol vapor, C2H5OH(g), at 83.5°C if 1.000 mol C2H5OH(g) occupies 41.00 L. Use the van der Waals equation. (See the table below for data.) Compare with the result from the ideal gas law.
Brian S.
Given the following information for ethanol, C2H5OH (at 1 atm), calculate the amount of heat in kJ needed (at 1 atm) to vaporize a 35.3-g sample of liquid ethanol at its normal boiling point of 78.4 °C. boiling point = 78.4 °C Hvap(78.4 °C) = 38.6 kJ/mol melting point = -115 °C Hfus(-115 °C) = 5.02 kJ/mol specific heat liquid = 2.46 J/g°C
Dj T.
The density of a gas is 1.96 g/L at 1.00 atm and 0 °C. What is the density of this gas at 0.855 atm and 25.0 °C?
Ankur S.
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD