Calculate the mass of the following:
(i) 0.5moles of `O_(2)` gas (ii) 0.5moles of O atoms
(iii) `3.011xx10^(23)"atoms of O"` (iv) `6.022xx10^(23)"molecules of "O_(2)`
(Givem: Gram atomic mass of oxygen=16g)
Gram of molecular mass of oxygen `(O_(2))`=32g)

Explanation

(i) 0.5 mole of `O_(2)` gas
No. of moles `= ("Mass of "O_(2) " in grams")/("Gram molecular mass")= (m)/(M)`
`therefore` Mass of `O_(2)` is gram (m) = No. of moles `xx M 0.5 xx (32g) = 16g`
(ii) 0.5 moles of oxygen (O) atoms
`"No. of gram atoms"=("Mass of oxygen (O) is grams")/("Gram atomic mass")=(m)/(M)`
Mass of oxygen of (O) in grams (m) = No. of moles `xx M = 0.5 xx (16g(0 = 8 g`
(iii) `3.011 xx 10^(23)` atoms of oxgen (O)
Step I : Calculation of no. of gram atoms of oxygen
`"No.of gram atoms"=("No.of atoms of oxygen")/("Avogadro's no.of atoms")= (N)/(N_(0))`
`= (3.011 xx 10^(23) )/(6.022xx 10^(23)) = 0.5` gram atom
Step II : Calculation of mass of oxgyen (O) atoms
Mass of oxygen (O) atoms =Gram atomic mass of oxygen `xx` No.of gram atoms of oxygen
`= (16g) xx 0.5 = 8 g`
(iv) `6.022 xx 10^(23)` molecules of oxygen`(O_(2))`
Step I: Calculation of no.of gram moles of oxygen.
`"No.of gram moles"=("No.of atoms of oxygen")/("Avogadro's no.of atoms")= (N)/(N_(0))`
`= (6.022 xx 10^(23))/(6.022 xx 10^(23)) = 1` gram mol.
Step II: Calculation of mass of oxgyen `(O_(2))` molecules.
Mass of oxygen `(O_(2))` molecules = Gram molecular mass of oxygen `xx` No of gram moles of oxgyen
`= (32 g) xx 1 = 23 g`.