Question

32g of O₂ have the same number of molecules as in *



(a) 16 g of CO

(b) 28 g of N₂

(c) 14 g of N₂

(d) 1.0 g of H₂

​

Answer 1

Answer:

your correct answer is 16 g of CO

Answer 2

Given:

32g of O₂

To find:

32g of O₂ have the same number of molecules as in the given options

Solution:

We know that,

$\boxed{\bold{Avogadro's\:Number, N_A \rightarrow 1 \:mole\: contains\: 6.022 \times 10^2^3 \:molecules}}$

and

$\boxed{\bold{No.\:of\:molecules\:= [no.\:of\:moles]\times [N_A]}}$

Finding the no. of molecules in 32 g of O₂:

No. of moles of O₂, n = $\frac{mass }{molar\: mass} = \frac{32}{32} = 1\: mole$

∴ No. of molecules in 32 g of O₂ is,

= n × $N_A$

= $1 \times 6.022\times 10^2^3$

= $6.022\times 10^2^3\:molecules$

Finding the no. of molecules in 16 g of CO:

No. of moles of CO, n = $\frac{mass }{molar\: mass} = \frac{16}{28} = 0.571 \:moles$

∴ No. of molecules in 16 g of CO is,

= n × $N_A$

= $0.571 \times 6.022\times 10^2^3$

= $3.438\times 10^2^3\:molecules$

Finding the no. of molecules in 28 g of N₂:

No. of moles of N₂, n = $\frac{mass }{molar\: mass} = \frac{28}{28} = 1 \:mole$

∴ No. of molecules in 28 g of N₂ is,

= n × $N_A$

= $1 \times 6.022\times 10^2^3$

= $6.022\times 10^2^3\:molecules$

Finding the no. of molecules in 14 g of N₂:

No. of moles of N₂, n = $\frac{mass }{molar\: mass} = \frac{14}{28} = 0.5 \:moles$

∴ No. of molecules in 14 g of N₂ is,

= n × $N_A$

= $0.5 \times 6.022\times 10^2^3$

= $3.011\times 10^2^3\:molecules$

Finding the no. of molecules in 1 g of H₂:

No. of moles of H₂, n = $\frac{mass }{molar\: mass} = \frac{1}{2} = 0.5 \:moles$

∴ No. of molecules in 1 g of H₂ is,

= n × $N_A$

= $0.5 \times 6.022\times 10^2^3$

= $3.011\times 10^2^3\:molecules$

From the above calculations, we conclude that:

32g of O₂ have the same number of molecules as in 28 g of N₂

⇒ option (b) is the correct answer

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