Answer:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Explanation:
By ideal gas equation:
[tex]PV=nRT[/tex]
Number of moles (n)
can be written as: [tex]n=\frac{m}{M}[/tex]
where, m = given mass
M = molar mass
[tex]PV=\frac{m}{M}RT\\\\PM=\frac{m}{V}RT[/tex]
where,
[tex]\frac{m}{V}=d[/tex] which is known as density of the gas
The relation becomes:
[tex]PM=dRT[/tex] .....(1)
We are given:
M = molar mass of chloroform= 119.5 g/mol
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]298K[/tex]
P = pressure of the gas = 1.00 atm
Putting values in equation 1, we get:
[tex]1.00atm\times 119.5g/mol=d\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\d=4.88g/L[/tex]
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
