Question

The density of acetone is 0.7899 g•cm-3. If 8.0 mL of acetone is vaporized at 81.0 oC and atmospheric pressure, what would the volume be in liters? (Do not include the units in your answer. Report a numerical answer only with significant digits. Assume 2 significant digits in atmospheric pressure.)

Answer 1

volume of liquid acetone = 8.0 mL
density = 0.7899 g/cm^3 = 0.7899 g/mL
So,
mass = density * volume
= 0.7899 g/mL * 8.0 mL
= 6.3192 g

acetone is CH3COCH3 or C3H6O

Molar mass of C3H6O,
MM = 3*MM(C) + 6*MM(H) + 1*MM(O)
= 3*12.01 + 6*1.008 + 1*16.0
= 58.078 g/mol


mass(C3H6O)= 6.3192 g

number of mol of C3H6O,
n = mass of C3H6O/molar mass of C3H6O
=(6.3192 g)/(58.078 g/mol)
= 0.1088 mol

Given:
P = 1.0 atm
n = 0.1088 mol
T = 81.0 oC
= (81.0+273) K
= 354 K

use:
P * V = n*R*T
1 atm * V = 0.1088 mol* 0.0821 atm.L/mol.K * 354 K
V = 3.16 L

Answer: 3.2 L