the density of acetone is 0.7899 g/cm-3. If 2mL of acetone is vaporized at 84 C and atmospheric pressure, what would the volume be in liters?
mass acetone = volume x density = ?
mols acetone = grams/molar mass = ?
Then PV = nRT and solve for V.
0.0489 L
aaron's an idiot the answer is 0.7971
To determine the volume of acetone in liters after vaporization, we need to convert the given volume of 2 mL to the desired unit.
1. Start with the given volume: 2 mL
2. Convert mL to cm³: Since 1 mL is equal to 1 cm³, the volume remains the same: 2 cm³
Now, we can proceed to solve the problem:
3. Given the density of acetone is 0.7899 g/cm³, we can calculate the mass of 2 cm³ of acetone using the formula: Mass = Volume × Density
Mass = 2 cm³ × 0.7899 g/cm³ = 1.5798 g
4. To convert the mass of acetone to moles, we need to know the molar mass of acetone. The molar mass of acetone (C₃H₆O) is approximately 58.08 g/mol.
Number of moles = Mass / Molar mass
Number of moles = 1.5798 g / 58.08 g/mol ≈ 0.0272 mol
5. Using the Ideal Gas Law (PV = nRT), we can calculate the volume of the vaporized acetone at 84 °C and atmospheric pressure. At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.
Volume = (Number of moles × Gas constant × Temperature) / Pressure
Volume = (0.0272 mol × 0.0821 L·atm/mol·K × (84 + 273.15) K) / 1 atm
Volume ≈ 0.717 L
Therefore, after vaporization, the volume of 2 mL of acetone at 84 °C and atmospheric pressure would be approximately 0.717 liters.