Question

(i)How much negative charge and how much positive charge are there on the electrons and the protons in a cup of water (0.25 kg)? Note Avogadro’s number is NA = 6.022 × 1023, and each oxygen atom has 8 electrons. (ii) What is the magnitude of the attractive force exerted by the electrons in a cup of water on the protons in a second cup of water at a distance of 10 m?

Answer 1

0.25 kg *(1e3 g/1kg)*(1 mol water/ 18g) = 13.88889 moles water

1 mol of water = 1 mol Oxygen and 2 mol Hydrogen

1 mol Hydrogen has 1 mol of proton and 1 mol of e-

And

1 mol Oxygen has 8 mol of proton and 8 mol of e-

13.88889 mol water *(1mol O/1mol water)*(8 mol e-/1mol O) = 111.111mol e-

13.88889 mol water *(2 mol O/1mol water)*(1 mol e-/1mol H) =27.77778 mol e-

so 138.8889 mol e- = 138.8889 mol e- *(-96485 C/1 mol e-) = -1.34 e7 Coulombs.

Since water is neutral. There is also 1.34e 7 Coloumbs ofprotons

F = k*q*q/r^2 = (8.898 e 9 N/m2/C2)*(1.34 e7 C)*(-1.34 e7 C)/(10m)^2 = 1.598 e 22 N

Note that both water cups are electrically neutral. So really thereis no net force (there are other coulombic interactions between theprotons and protons of both cups of water. Also there iselectron-electron repulsion. The sum of the forces is zero.) Butstill, the above number is the force of attraction between only theelectrons and protons of individual cups.