The maximum number of molecules is present in :

Options

A. 15 L of `H_(2)` gas at STP
B. 5 L of `N_(2)` gas at STP
C. 0.5 g of `H_(2)` gas
D. 10 g of `O_(2)` gas

Answer

A

Explanation

In 15 L of `H_(2)` gas at STP,
the number of molecules `=(6.023xx10^(23))/(22.4)xx15`
`=4033xx10^(23)`
In 5 L of `N_(2)` gas at STP, ltbr. the number of molecules `=(6.023xx10^(23)xx5)/(22.4)`
`=1.344xx10^(23)`
In 0.5 g of `H_(2)` gas,
the number of molecules `= (6.23xx10^(23)xx0.5)/(2)`
`=1.505xx10^(23)`
In 10 g of `O_(2)` gas ,
the number of molecules `=(6.23xx10^(23)xx10)/(32)`
`=1.882xx10^(23)`
Hence , maximum number of molecules are presents in 15 L of `H_(2)` at STP.