Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

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Solution

Molar mass of dinitrogen $(N_{2}) = 28 g m o l^{- 1}$
Thus, $1.4 g o f N_{2} = \frac{1.4}{28} = 0.05 m o l$
$= 0.05 \times 6.02 \times 10^{23}$ number of molecules
$= 3.01 \times 10^{23}$ number of molecules
Now, 1 molecule of $N_{2}$ contains 14 electrons
Therefore, $3.01 \times 10^{23}$ molecules of $N_{2}$ contains $= 14 \times 3.01 \times 10^{23}$
$= 4.214 \times 10^{23}$ electrons

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Stoichiometric Calculations

Standard XII Chemistry

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Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Molar mass of dinitrogen (N2)=28gmol−1 Thus, 1.4 g of N2=1.428=0.05 mol =0.05×6.02×1023 number of molecules =3.01×1023 number of molecules Now, 1 molecule of N2 contains 14 electrons Therefore, 3.01×1023 molecules of N2 contains =14×3.01×1023 =4.214×1023 electrons