Question

Two bulbs B1 and B2 of equal capacity contain 10g oxygen ($O_{2}$) and ozone ($O_{3}$) respectively. Which bulb will have greater number of O-atoms and which will have greater number of molecules?

Answer 1

1 mole of molecules contains $6.022 \times { 10 }^{ 23 } molecules.$

The given gas is oxygen.

Atomic number of oxygen is 8

Molar mass of ${ O }_{ 2 }$ gas is 32 g/mol

Molar mass of ${ O }_{ 3 }$ gas is 48 g/mol

For flask A:

It contains 10 g of oxygen

$Number\quad of\quad oxygen\quad molecules = 6.022 \times { 10 }^{ 23 } \times \frac { 10 }{ 32 } = 1.88 \times {10 }^{ 23 }$

$Number\quad of\quad atoms = 2 \times 1.88 \times { 10 }^{ 23 } = 3.76 \times  { 10 }^{ 23 }$

For B flask:

It contains 10 g of ozone

$Number\quad of\quad ozone\quad molecules = 6.022 \times { 10 }^{ 23 } \times \frac { 10 }{ 48 } = 1.254 \times { 10 }^{ 23 }$

$Number\quad of\quad atoms = 3 \times 1.254 \times { 10 }^{ 23 } = 3.76 \times { 10 }^{ 23 }$

Therefore, flask A has more molecules compared to flask B while both the flasks has same number of oxygen atoms.

Answer 2

Answer:

10 g O

2

=

32

10

mole =

32

10

×6.02×10

23

molecules

=1.88×10

23

molecules =2×1.88×10

23

atoms =3.76×10

23

atoms

10 g O

3

=

48

10

mole =

48

10

×6.02×10

23

molecules

=1.254×10

23

molecules =3×1.254×10

23

atoms =3.76×10

23

atoms

This both contain the same number of atoms but bulb B

1

contains more number of molecules