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.)

Respuesta :

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 = [tex]263 cm^{3}[/tex] = 263 mL

So, weight of water = 263 g

Number of moles of water = [tex]\frac{Given mass in g}{molar mass}[/tex]

= [tex]\frac{263 g}{18 g/mol}[/tex]

= 14.61 g/mol

Number of protons = [tex]14.61 g/mol\times 10\times 6.023 \times 10^{23}[/tex]

=  [tex]87.99603 \times 10^{24}[/tex]

As, charge on one proton = [tex]1.6\times 10^{-19} C[/tex]

Thus, total charge = [tex]1.6\times 10^{-19} C\times 87.99603 \times 10^{24}[/tex]

= [tex]140.793 \times 10^{5} C[/tex]

=  [tex]1.40 \times 10^{7} C[/tex]

Thus, [tex]1.40 \times 10^{7} C[/tex] is the total number of coulombs of positive charge.