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In which case is the number of molecules of water maximum?A. 18 mL of waterB. 0.18g of waterC. 0.00224 L of water vapours at 1 atm and 273 KD. `10^(-3)` mol of water |
Answer» Correct Answer - A (1) 18 mL water As `d_(H_(2)O)=1g//mL" "`So `W_(H_(2)O)=18 g` `n_(H_(2)O)=18/18=1` molecules `=1xxN_(A)` (2) 0.18 g of water `n_(H_(2)O)=0.18/18=0.01` (molecules) `H_(2)O=0.01xxN_(A)` (3) `(V_(H_(2)O(g)))_(STP)=0.00224 L` `n_(H_(2)O)=V/22.4=0.00224/22.4=0.0001` Molecules `=0.0001xxN_(A)` (4) `n_(H_(2)O)=10^(-3)` `("molecules")_(H_(2)O)=10^(-3)xxN_(A)` |
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