Saved Bookmarks
| 1. |
Calculate the number of molecules present (i) in 34.20 grams of cane sugar (C_(12)H_(22)O_(11)) (ii) in one litre of water assuming that the density of water is 1g//cm^(3). (iii) in one drop of water having mass 0.05 g. |
|
Answer» Solution :1 mole of `C_(12H_(22)O_(11)=342g` `[because" Molecular mass of CANE sugar "(C_(12)H_(22)O_(11))=12xx12+22xx1+11xx16=342" amu"]` `=6.022xx10^(23)" MOLECULES"` Now, 342 g of cane sugar contain `6.022xx10^(23)" molecules"` `therefore"34.2g of cane sugar will contain"=(6.022xx10^(22))/(342)xx34.2=6.022xx10^(23)" molecules"` (ii) 1 mole of water = 18 g = `6.022xx10^(23)" molecules"` Mass of 1 litre of water = `"Volume"XX"density"=100mLxx1"g mL"^(-1)=1000g` `"Now,18 g of water contain"=6.022xx10^(23)" molecules"` `therefore"1000 g of water will contain"=(6.022xx10^(23)xx1000)/(18)` `=3.346xx10^(25)" molecules"` (iii) 1 mole of `H_(2)O` = 18 g = `6.022xx10^(23)" molecules"` Mass of 1 drop of water = 0.05 g Now, 18 g of `H_(2)` contain = `6.022xx10^(23)" molecules"` `therefore 0.05g" of "H_(2)O" will contain"=(6.022xx10^(23))/(18)xx0.05=1.673xx10^(21)" molecules"` |
|