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.

Explanation

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"`