What is the mass of each one of the following ?
(a) `1` mole of water `(H_(2)O)` , (b) `1` mole of ethanol `(C_(2)H_(6)O)`
(c) `1` mole of glucose `(C_(6)H_(12)O_(6))` , (d) `1` mole of cane sugar `(C_(12)H_(22)O_(11))`
(Atomic masses : ` H = 1 u , C = 12 u , O = 16 u`)

Explanation

The mass of 1 mole of each one of the given substances will be equal to their respective molecular masses expressed in 'grams'. Now :
(a) 1 mole of `H_(2)O` = Molecular mass of `H_(2)O` in grams
`=` Mass of `H xx 2 +` Mass of O
`= 1 xx 2 + 6`
`= 2 + 16`
`= 18` grams
So, the mass of 1 mole of water `(H_(2)O)` is `18 g`.
(b) 1 mole of `C_(2)H_(6)O` = Molecular mass of `C_(2)H_(6)O` in grams
= Mass of `C xx 2` + Mass of `H xx 6 `+ Mass of O
`= 12 xx 2 + 1 xx 6 + 16`
`= 24 + 6 + 16`
`= 46` grams
So, the mass of 1 mole of ethanol `(C_(2)H_(6)O)` is `46 g`.
(c ) 1 mole of `C_(6)H_(12)O_(6)` = Molecular mass of `C_(6)H_(12)O_(6)` in grams
= Mass of `C xx 6` + Mass of `H xx 12` + Mass of `O xx 6`
`= 12 xx 6 + 1 xx 12 + 16 xx 6`
`= 72 + 12 + 96`
`= 180` grams
So, the mass of 1 mole of glucose `(C_(6)H_(12)O_(6))` is `180 g`
(d) 1 mole of `C_(12)H_(22)O_(11)` = Molecular mass of `C_(12)H_(22)O_(11)` in grams
`=` Mass of `C xx 12` + Mass of `H xx 22` + Mass of `O xx 11`
`= 12 xx 12 xx +1 xx 22 + 16 xx 11`
`= 342` grams
So, the mass of 1 mole of cane sugar `(C_(12)H_(22)O_(11))` is `342 g`.