Question

the gram molecules in 21 gram of Nitrogen gas

Answer 1

Explanation:

molar mass of nitrogen gas (N2)=28 g

given mass of N2=21 g

moles of N2

=given mass of N2/molar mass of nitrogen gas (N2)

=21/28

=0.75moles

molecules of N2=0.75 NA (NA represent Avogadro's no)

=0.75*6.022*10²³

=4.5165*10²³ gram molecules

Answer 2

The gram molecules in 21 g of nitrogen gas is $4.517\times 10^{23}$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{21g}{28g/mol}=0.75moles$

1 mole of nitrogen $(N_2)$ =$6.023\times 10^{23}$ g molecules of nitrogen $(N_2)$

0.75 mole of nitrogen $(N_2)$ = $\frac{6.023\times 10^{23}}{1}\times 0.75=4.517\times 10^{23}$ g molecules of nitrogen $(N_2)$

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