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Solution:

$19) A m o u n t o f o x y g e n i n 1 l i t r e a i r = 80 % = \frac{80}{100} \times 1000 = 800 m l 1 m o l e o f a n y g a s a t S T P o c c u p i e s 22.4 l i t r e s 22400 m l o f O_{2} g a s c o n t a i n s 6.022 \times 10^{23} m o l e c u l e s 1 m l o f o x y g e n g a s c o n t a i n s = \frac{6.022 \times 10^{23}}{22400} 800 m l c o n t a i n s = \frac{6.022 \times 10^{23}}{22400} \times 800 = 2.15 \times 10^{22} m o l e c u l e s M o l e c u l e s p r e s e n t i n 1 l i t r e a i r c o n t a i n i n g 80 % v o l u m e a t S T P i s 2.15 \times 10^{22} m o l e c u l e s$

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