The system that contains the maximum number of atoms is

Options

A. `4.25` g of `NH_(3)`
B. 8 g of `O_(2)`
C. 2g of `H_(2)`
D. 4 g of He

Answer

C

Explanation

Gram molecular mass of `NH_(3)` is 17 g
`therefore` No. of molecules in 4.25 g of `NH_(3)`
`=4.25/18 N_(A) = N_(A)/4`
Now, one molecule of `NH_(3)` contains 4 atoms.
`therefore N_(A)/4` molecules contain `N_(A)/4 xx 4 = N_(A)` atoms
Again, `32 g` of `O_(2) = N_(A)` molecules = `2N_(A)` atoms
`therefore 8 g` of `O_(2) = N_(A)/32 xx 8 = N_(A)/4` molecules
`=(2N_(A))/32 xx 8 = N_(A)/2` atoms
On the other hand,
2g of `H_(2) = N_(A)` molecules = `2N_(A)` atoms
4g of He = `N_(A)` atoms [`therefore` gram atomic mass of He = 4g]