The maximum number of molecules are present in

Options

A. `15L of H_2gas at STP`
B. `5 L of N_2 gas at STP`
C. `0.5 g of H_2gas`
D. `10 g of O_2 gas`

Answer

A

Explanation

(a) No. of molecules in different cases (a) `therefore` 22.4 liter at STP contains `=6.023xx10^(23)`molecule of `H_2` `therefore` 15 liter at STP contains`=(15)/(22.4)xx6.023xx10^(23)=4.03xx10^(23)`molecule of `H_2` (b) `therefore`22.4 liter at STP contains `=6.023xx10^(23)`molecules of `N_2` `therefore` 5 liter at STP contains`=5/(22.4)xx6.023xx10^(23)=1.344xx10^(23)`molecule of `N_2` (c ) `therefore 2gm of H_2-6.023xx10^(23)`molecule of `H_2 therefore 0.5 gm of H_2=0.5/2xx6.023xx10^(23)=1.505xx10^(23)`molecule of `H_2` (d) Similarly 10 g of `O_2` gas `10/32xx6.023xx10^(23)`molecules of `O_2=1.88xx10^(23)`molecule of `O_2`. Thus (a) will have maximum number of molecules.