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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)] |
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Answer» Solution :(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` |
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