(a) A body is having charge `+4.8 xx 10^(-18) C`. Estimate the number of electrons in excess/shortage.
(b) A body is having charge `3 xx 10^(-12) C`. Determine whether this statement is true/false.
(c) Estimate the number of electrons in `36 g` of water. How much is the total negative charge on these electrons.
(d) Estimate the charge in `26 g` of `Na^(+)`.

Explanation

`q = n e`
`n = (q)/(e) = (4.8 xx 10^(-18))/(1.6 xx 10^(-19)) = 30`
Since charge is positive i.e. shortage of electrons.
(b) `n = (q)/( e) = ( 3 xx 10^(-12))/(1.6 xx 10^(-19)) = 1.875 xx 10^(7)`
`= 1875 xx 10^(4) : a` whole number
The charge on a body should be some integral multiple of `e`. Here `n` is a whole number i.e. body can have this charge , true,
( c) Number of moles `= (36)/(18) = 2`
Number of molecules `= 2 xx` Avogadro's number
`= 2 xx 6.023 xx 10^(23)`
Since one molecule of water `(H_(2) O)` is having `10` electrons.
Number of electrons `= 2 xx 6.023 xx 10^(23) xx 10`
`= 12 xx 10^(24)`
Total negative charge on electrons `= 12 xx 10^(24) xx 1.6 xx 10^(-19)`
`= 1.92 xx 10^(6) C`
Water is neutral.
Positive charge = Negative charge
Net charge `= 0`
(d) Number of atoms ` = (26)/(13) = 2`
Number of atoms ` = 2 xx 6.023 xx 10^(23)`
One atom is having charge `+e`
Total charge `= 2 xx 6.023 xx 10^(23) xx 1.6 xx 10^(-19)`
` = 19.27 xx 10^(4) C`