Question

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

Answer 1

Answer:

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

Explanation:

By ideal gas equation:

$PV=nRT$

Number of moles (n)

can be written as: $n=\frac{m}{M}$

where, m = given mass

M = molar mass

$PV=\frac{m}{M}RT\\\\PM=\frac{m}{V}RT$

where,

$\frac{m}{V}=d$ 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:

$1.00atm\times 119.5g/mol=d\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\d=4.88g/L$

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