Question

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

Answer 1

Answer:

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

Explanation:

$N=n\times N_A$

Where:

N = Number of particles / atoms/ molecules

n = Number of moles

$N_A=6.022\times 10^{23} mol^{-1}$ = Avogadro's number

We have:

N = $1.5\times 10^{24}$

n =?

$n=\frac{N}{N_A}=\frac{1.5\times 10^{24}}{6.022\times 10^{23} mol^{-1}}$

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:

$V=\frac{2.4909 mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}$

V = 55.86 L

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