1.

16 of oxygen has same number of molecules as inA. 16 g of COB. 28 g of `N_(2)`C. `14 g "of" N_(2)`D. `1.0g "of" H_(2)`

Answer» The number of molecuels can be calculated as follows
Number of molecuels `=("Mass")/(" Molar mass")xx ` Avogadro number `(N_(A))`
Number of molecuels, in 16 g of oxygen `=(16)/(32)xxN_(A)=(N_(A))/(2)`
In `16g` of `CO=(16)/(28)xxN_(A)=(N_(A))/(175)`
In 28 g of `N_(2)=(14)/(28)xxN_(A)=(N_(A))/(2)`
In `1 g "of " H_(2)=(1)/(2)xxN_(A)=(N_(A))/(2)`
So, `16 g "of" O_(2)=14g "of" N_(2)=1.0g "of" H_(2)`


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