Question

What is the density of CHCL3 vapor at 1.00atm and 298K?

Answer 1

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

Step-by-step explanation:

By ideal gas equation:

`PV=nRT`

Number of moles (n)

can be written as:
`n=(m)/(M)`

where, m = given mass

M = molar mass

`PV=(m)/(M)RT\\\\PM=(m)/(V)RT`

where,

`(m)/(V)=d` which is known as density of the gas

The relation becomes:

`PM=dRT` .....(1)

We are given:

M = molar mass of chloroform= 119.5 g/mol

R = Gas constant =
`0.0821\text{ L atm }mol^(-1)K^(-1)`

T = temperature of the gas =
`298K`

P = pressure of the gas = 1.00 atm

Putting values in equation 1, we get:

`1.00atm* 119.5g/mol=d* 0.0821\text{ L atm }mol^(-1)K^(-1)* 298K\\\\d=4.88g/L`

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.