Question

the gram molecules in 21 gram of Nitrogen

Answer 1

 Molecular mass of N2 is 14.0 + 14.0 = 28.0 g/mole 
One moles contains Avogadro's number of molecules:6.02 x 10^23 molecules/mole 

(21 g N2)(1 mole/28.0 g)(6.02 x 10^23 molecules/mole) =

Answer 2

The number of gram molecules of given amount of nitrogen gas are $4.52\times 10^{23}$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of nitrogen gas = 21 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

$\text{Moles of nitrogen gas}=\frac{21g}{28g/mol}=0.75mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of gram molecules

So, 0.75 moles of nitrogen gas will contain = $(0.75\times 6.022\times 10^{23}=4.52\times 10^{23}$ number of gram molecules

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