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**

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