The system that contains the maximum no. of atoms is-

Options

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

Answer

C

Explanation

No. off molecules in 4.25g of `NH_(3)=(4.25)/(17)N_(A)=(N_(A))/(4)`
Now, one molecule of `NH_(3)` contains 4 atoms
`therefore (N_(A))/(4)` molecules contain `(N_(A))/(4)xx4=N_(A)` atoms.
Again, 32g of `O_(2)=N_(A)` molecules=`2N_(A)` atoms
`therefore`8g of `O_(2)=(N_(A))/(32)xx8=(N_(A))/(4)` molecules
`=(2N_(A))/(32)xx8` atoms`=(N_(A))/(2)` atoms.