How many molecules are present in 7 grams of n2

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Solution

Dear student,

$m o l e c u l a r m a s s o f n i t r o g e n (N 2) = 28 g 1 m o l e o f n i t r o g e n c o n t a i n s 28 g o f N 2 w i l l b e h a v i n g 6.023 \times 10^{23} m o l e c u l e s S o 1 g o f N 2 w i l l c o n t a i n (6.023 \times 10^{23}) \div 28 H e n c e 7 g o f N 2 c o n t a i n \frac{(6.023 \times 10^{23})}{28} \times 7 = 1.5 \times 10^{23} m o l e c u l e s$

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