Maximum number of molecules are present in

Options

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

Answer

A

Explanation

(a) 15 L of `H_(2)` gas at S.T.P
`= (15L)/((22.4L))xx6.022xx10^(23)=4.033xx10^(23)`
(b) 5L of `N_(2)` gas at S.T.P.
`= ((5L))/((22.4L))xx6.022xx10^(23)`
`=1.344xx10^(23)`
(c) 0.5 g of `H_(2)=((0.5g))/((2.0g))xx6.022 xx 10^(23)`
`=1.505 xx 10^(23)`
10 g of `O_(2)=((10.0g))/((32.0g))xx6.022xx10^(23)`
`=1.882xx10^(23)`.