The maximum number of molecules is present in

Options

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

Answer

B

Explanation

More the volume, more is the molecules. Converted the choices (A) and (D) into volumes.
(A) 32 g of `O_(2)` at S.T.P = 22.4 L
`:.` 10 g `O_(2)` at S.T.P. `= (22.4)/(32)xx10L=7L`
(Molar mass of `O_(2) = 32 g mol^(-1)`)
(D) 2 g of `H_(2)` at S.T.P. = 22.4 L
0.5 g of `H_(2)` at S.T.P. `= (22.4)/(2)xx0.5L=5.6L`
Thus we have 7L of `O_(2), 15 L` of `H_(2)`, 5 L of `N_(2)` and 5.6 L of `H_(2)` (at S.T.P.)
Clearly 15 L of `H_(2)` has maximum number of molecules