Question

Which has a greater number of atoms? (a) 48 g of O2 or (b) 22 g of CO2.​

Answer 1

Given :-

• Which has a greater number of atoms?

        a) 48g of O₂    (or)      b) 22g of CO₂

Solution :-

• First of all, we know that the number of atoms can be calculated using the formula :-

        No.of atoms = no.of moles(n) × Avogadro's number(N) × Atomicity

       ⇒ Here, 1) number of moles is Weight/Gram molecular weight

                      2) Atomicity is number of atoms in one molecule of element

                      3) Avogadro's number is 6.023 × 10²³

• Now let us first find no.of atoms in 48g of O₂

     → Number of atoms in 48g of O₂ =  48/32  × N × 2  = 3/4 × N = 0.75N

• Now let's find no.of atoms in 22g of CO₂

    → Number of atoms in 22g of CO₂ = 22/44 × N × 3 = 3/2 × N  = 1.5N

⇒ Clearly, 1.5N > 0.75N

∴ 22g of CO₂ has greater no.of atoms!

Thank you, please mark as brainliest!

Answer 2

Answer:

48 g of $O_{2}$ has a greater number of molecules.

Explanation:

(a)  48 g of $O_{2}$

We know that the molar mass of $O_{2}$  is 32 g/mol.

Therefore the number of moles = given mass/molar mass

                                                    = $\frac{48}{32}$ = 1.5 moles

One mole of compound contains an Avogadro number of particles.

Therefore the number of molecules of oxygen molecules in 1.5 moles

                               = 6.023 × $10^{23}$ × 1.5 = 9.0345 × $10^{23}$

(b) 22 g of $CO_{2}$

We know that the molar mass of $CO_{2}$ is 44 g/mol

Therefore the number of moles = given mass/molar mass

                                                = $\frac{22}{44}$ = 0.5 moles

One mole of compound contains an Avogadro number of particles.

Therefore the number of molecules of carbon dioxide molecules in 0.5 moles  = 6.023 × $10^{23}$ × 0.5 = 3.0115 × $10^{23}$