Calculate the number of g-molecules (mole of molecules) in the following : (i) 3.2 gm `CH_(4)` (ii) 70 gm nitrogen (iii) `4.5xx10^(24)` molecules of ozone (iv) `2.4xx10^(21)` atoms of hydrogen (v) 11.2 L ideal gas at `0^(@)` C and 1 atm (vi) 4.54 ml `SO_(3)` gas at STP (vii) 8.21 `L C_(2)H_(6)` gas at 400 K and 2 atm (viii) 164.2 ml He gas at `27^(@)C` and 570 torr `[N_(A)=6xx10^(23)]`

Explanation

(i) 3.2 gram `CH_(4)`
number of moles `(CH_(4))=(w)/(M)=(3.2)/(16)=0.2` moles
(ii) 70 gram `N_(2)`
Number of moles `=(w)/(M)=(70)/(28)=2.5`
(iii) `4.5xx10^(24)` molecules of `O_(3)`
Number of moles `=("no. of molecules ")/(N_(A))=(4.5xx10^(24))/(6xx10^(23))=7.5`
(iv) `2.4xx10^(21)` atoms of hydrogen
Number of gram molecules of `H_(2)=("no. of molecules")/(N_(A))=(2.4xx10^(21))/(2xx6xx10^(23))=0.002`
(v) 11.2 litre ideal gas at `0^(@)` C and 1 atm
Number of moles `=("Volume at " 0^(@) C & 1 atm)/(22.4 "litres")=(11.2)/(22.4)=0.5`
(vi) 4.54 mol `SO_(3)` gas at STP
Number of moles `=(V_("STP")(ml))/(22700ml)=(4.52)/(22700)=2xx10^(-4)`
(vii) 8.21 litre `C_(2)H_(6)` at 400 K and 2 litre
`=n(PV)/(R.T)=(2xx8.21)/(0.0821xx400)=0.5`
(viii) `164.2xx` ml He gas at `27^(@)C and 570` torr
`n=(PV)/(RT)=((570)/(760)atm)xx(164.2xx10^(-3) "litre")/(0.0821xx300)=0.005`