Question

Calculate the number of coulombs of positive charge in 263 cm3 of (neutral) water. (hint: a hydrogen atom contains one proton; an oxygen atom contains eight protons.)

Answer 1

Since, hydrogen atom consist of one proton and an oxygen atom consist of eight protons.

Now, density of water is 1 g/ml

Given volume =
`263 cm^(3)` = 263 mL

So, weight of water = 263 g

Number of moles of water =
`(Given mass in g)/(molar mass)`

=
`(263 g)/(18 g/mol)`

= 14.61 g/mol

Number of protons =
`14.61 g/mol* 10* 6.023 * 10^(23)`

=
`87.99603 * 10^(24)`

As, charge on one proton =
`1.6* 10^(-19) C`

Thus, total charge =
`1.6* 10^(-19) C* 87.99603 * 10^(24)`

=
`140.793 * 10^(5) C`

=
`1.40 * 10^(7) C`

Thus,
`1.40 * 10^(7) C` is the total number of coulombs of positive charge.