Forr 14 g of CO, the wrong statement is

Options

A. it occupies 2,24 L at NTP
B. it corresponds to `(1)/(2)` mole of CO
C. it corresponds to same mole of CO and nitrogen gas
D. it corresponds to `3.01xx10^(23)` molecules of CO

Answer

D

Explanation

(d). 28 g of CO occupies=22.4L ltbr of CO at NTP. Thus, 14 g of CO occupies 11.2 L volume at NTP.
(b) Number of moles of CO
`=("Mass of O")/("molar mass of CO")`
`=(14g)/("28 g "mol^(-1))=(1)/(2)" mol CO"`
Thus, 14 g CO equal to `(1)/(2)` mole of CO.
(c) As the molar mass of CO and `N_(2)` gas are same, thus 14 g of CO corresponds to same moles of CO and `N_(2)` gas.
(d) 28 g of CO contans=`6.022xx10^(23)`
molecules
`therefore14g` of CO contains
`=(6.022xx10^(23))/(28)xx14`
`=3.011xx10^(23)` molecules
thus, 14 g of CO corresponds to `3.011xx10^(23)` molecules.