Which of the following pairs will have the same number of atoms ?
[ atme wt. `H=1, O=16, C=12, Cl=35.5 g "mol"^(-1)`]

Options

A. 28 g CO and 36.5 g HCl
B. 44 g `CO_(2)` and 44 g CO
C. 14 g `N_(2)` and `28 g CO_(2)`
D. `28 g N_(2)` and 36.5 g HCl

Answer

A, B, C, D

Explanation

(A) (i) Molecular mass of CO `=(12+16)`
`= 28 g "mol"^(-1)`
`:.28 g CO = 1 ` mole CO
(ii) Molecular mass of HCL `=(1+35.5)`
`= 36.5 f "mol"^(-1)`
`:.36.5 g HCl = 1 ` mole HCl
As CO and HCl both are 1 mole and they have same 2 atoms, hence will have same number of atoms
(B) `44 g CO_(2) =1 "mole" CO_(2)`
but `44g CO != 1 ` mole CO
`:.` They do not have equal number of molecules
(C) `14 g N_(2)`
`= (14)/(28) "mole" N_(2)`
`= 1/2 "mole" N_(2)`
`28 g CO_(2)`
`= (28)/(44) ` mole `CO_(2)`
`= 0.63.63` mole `CO_(2)`
`:. 14 g N_(2)` and `28 g CO_(2)` have have different moles and as a result different no. of molecules
(D) `28 g N_(2)`
`= (28)/(28) ` mole `N_(2)`
` = 1` mole `N_(2)`
`=2` mole N atoms
`18 g H_(2)O`
`= (18)/(18) ` mole `H_(2)O`
` = 1` mole `H_(2)O`
` = 3` mole atoms
`:.N_(2)` and `H_(2)O` have different no. of atoms