(i) Calculate the mass of 3.011 x 1023 nitrogen molecules.
(ii) How many moles are 12.044 x 1023 atoms of He?

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Solution

$(i) M a s s o f n i t r o g e n m o l e c u l e s =$
$\frac{N u m b e r o f m o l e c u l e s}{A v o g a d r o n u m b e r} \times M o l a r m a s s M a s s (m) o f n i t r o g e n m o l e c u l e s = \frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} \times 28 = 14 g (i) N o o f M o l e s = \frac{N u m b e r o f m o l e c u l e s}{A v o g a d r o n u m b e r} = \frac{12.044 \times 10^{23}}{6.022 \times 10^{23}} = 2 m o l e o f a t o m$

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