Question

The vapor pressure of liquid acetone, CH3COCH3, is 100 mm Hg at 281 K. A 6.06E-2 g sample of liquid CH3COCH3 is placed in a closed, evacuated 360. mL container at a temperature of 281 K. Calculate what the ideal gas pressure would be in the container if all of the liquid acetone evaporated.

Answer 1

Step-by-step explanation:

According to ideal gas equation, product of pressure and volume equals the product of number of moles, gas constant and temperature.

Mathematically, PV = nRT

where P = pressure, V = volume

n = no. of moles, R = gas constant = 0.0821 atm L/mol

T = temperature

Since, it is known that number of moles equal mass divided by molar mass.

Hence, number of moles of given sample of acetone are as follows.

No. of moles =
`\frac{mass}{\text{molar mass of acetone}}`

=
`(6.06 * 10^(-2))/(58 g/mol)`

=
`0.104 * 10^(-2)` mole

Therefore, putting the values in ideal gas equation as follows.

PV = nRT

`P * 0.36 L = 0.104 * 10^(-2) * 0.082 atm L/mol * 281 K`

= 0.359 atm

In 1 atm equal to 760 mm Hg. So, convert 0.359 atm into mm Hg as follows.

`0.359 atm * (760 mm Hg)/(1 atm)`

= 272.84 mm Hg

Hence, pressure of the ideal gas will be 100 mm Hg + 272.84 mm Hg = 372.84 mm Hg

Thus, we can conclude that the ideal gas pressure in the container if all of the liquid acetone evaporated is 372.84 mm Hg.