How many molecules of oxygen would occupy a volume of 2.24 cm\(^{3}\) at s.t.p?

[Molar volume at s.t.p = 22,400 cm\(^{3}\), Avogadro's number = 6.02 x 10\(^{23}\)]

Kennykhay

3 years ago

first formular to consider :mole=volume/22.4 dm^3 at stp
therefore we convert the volume and molar volume from cm^3 to dm^3 that's divided by 1000
.
. . mole = 0.0024/22.4 =0.0001

the second formula is mole=number of molecules/6.02*10^23

.
. .0.0001=number of moles/6.02*10^23

1*10^-4=no. of molecules/6.02*10^23

Cross multiply ,,,u have
1*10^-4 *6.02*10^23 =6.02 *1 * 10^23-4

answer=6.02*10^19

Option C is d correct answer

Award0001

2 years ago

simple approach
since 1 mole of a gas occupies a volume of 22400cm³ and contains 6.02×10^23 atoms.
That means that one mole of oxygen O2 occupies a volume of 22400cm³
22400cm³ contains 6.02×10^23 atoms
2.24cm³ will contain
2.24/22400 × 6.02×10^23 = 0.000602×10^23
same as 6.02×10^19