Question

What volume does 1.5 x 10^24 molecules of CO2 occupy at STP?

Answer 1

The volume occupied by given molecules of carbon dioxide at STP is 55.86 L.

Step-by-step explanation:

`N=n* N_A`

Where:

N = Number of particles / atoms/ molecules

n = Number of moles

`N_A=6.022* 10^(23) mol^(-1)` = Avogadro's number

We have:

N =
`1.5* 10^(24)`

n =?

`n=(N)/(N_A)=(1.5* 10^(24))/(6.022* 10^(23) mol^(-1))`

n = 2.4909 moles

Using ideal gas equation:

PV = nRT

where,

P = Pressure of carbon dioxide gas =
`1 atm` (at STP)

V = Volume of carbon dioxide gas = ?

n = number of moles of carbon dioxide gas =2.4909 mol

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of carbon dioxide gas = 273.15 K (at STP)

Putting values in above equation, we get:

`V=(2.4909 mol* 0.0821 atm L/mol K* 273.15 K)/(1 atm)`

V = 55.86 L

The volume occupied by given molecules of carbon dioxide at STP is 55.86 L.