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Calculate the volume of STP occupied by (i) 14 g of nitrogen, (ii) 1.5 moles of carbon dioxide and (iii) 10^(21) molecules of oxygen. |
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Answer» SOLUTION :(i) Molecular mass of nitrogen = 28 U 1 mole of nitrogen = 28 g = 22.4 L at STP, i.e., 28 g of nitrogen occupy 22.4 L at STP `therefore" 14 g of nitrogen will occupy"=(22.4)/(28)xx14=11.2"L at STP"` (ii) 1 mole of carbon dioxide have volume = 22.4 L at STP `therefore "1.5 MOLES of carbon dioxide will occupy volume"=(22.4)/(1)xx1.5L=33.6"L at STP"` (iii) 1 mole of `O_(2)` molecules = `6.022xx10^(23)" molecules"=22.4" L at STP"` i.e., `6.022xx10^(23)` molecules of oxygen will occupy 22.4 L at STP `10^(21)` molecules of oxygen will occupy `=(22.4)/(6.022xx10^(23))xx10^(21)"L at STP"` `=3.72xx10^(-2)"L at STP"=3.72xx10^(-2)xx10^(3)"mL at STP"=37.2"mL at STP"` |
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