1.

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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