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If 1 g of each of the following gases are taken at STP, which of the gases will occupy (a) greatest volume and (b) smallest volume ?CO, H_(2)O, CH_(4), NO |
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Answer» SOLUTION :From Avogardro.s law, we know that Bolume of 1 mole of GAS = gram molecular mass = 22.4 L at STP. Volume of occupied by 28 g CO (1 MOL CO) = 22.4 L at STP. (`because` Molar mass of `CO=12+16=28 g mol^(-1)`) `terefore` Volume occupied by 1 g `CO=(22.4)/(28)` L at STP. Similarly, Volume occupied by 1 g `H_(2)O=(22.4)/(18)` L at STP.(`because` Molar mass of `H_(2)O = (2xx1)+16)=18 g mol^(-1)`) Volume occupied by 1 g `CH_(4)=(22.4)/(16)` L at STP.(`because` Molar mass of `CH_(4)=12+(4xx1)=16 g mol^(-1)`) Volume occupied by 1 g NO `= (22.4)/(30)` L at STP.(`because` Molar mass of NO `= 14+16 = 30 g mol^(-1)`) Hence, 1 g `CH_(4)` will occupy maximum volume and 1 g of NO will occupy minimum volume at STP. |
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