What is the Number of Coulombs of Positive Charge in Water?

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To calculate the number of Coulombs of positive charge in 292 cm³ of water, one must consider the number of protons in a water molecule, which totals 10 (2 from hydrogen and 8 from oxygen). The calculation involves determining the number of moles of water in the given volume, then converting that to the number of molecules, and finally multiplying by the charge of a proton. The initial calculation mistakenly used 18 protons instead of the correct total of 10, leading to an incorrect result. Understanding the dissociation of water into hydronium and hydroxide ions may also be relevant for a more accurate assessment.
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Here's the problem:

Calculate the number of Coulombs of positive charge in 292 cm3 water.

I don't need it done for me or anything. I'm sure I've missed something simple, but if someone can give me a hint on how to get started, I think I can solve this one.
 
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Hi there Gen and welcome to PF,

Note that there should be no net charge. However, if you wish to calculate the amount of positive charge present I'll start you off by asking you some questions. How many protons in a water molecule? How many water molecules in a mole of water? How many protons is a mole of water? How many moles of water in 292cm3? What is the charge in coulombs of a proton?
 
I got the wrong answer somehow. Your hints were very helpful. Here is what I did:

(1 mol Water/18 grams Water) * (1 gram Water/1 cm3 Water) * (292 cm3 Water) * (6.022E23 molecules Water/1 mol Water) * (18 protons Water/1 molecule Water) * (1.602E-19 Coulombs/1 proton Water) giving me a result of 2.817E7 coloumbs, but the online grading thing says that's wrong. Did I mess something up?
 
Your math looks Ok. Can the question possibly relate to the presence of hydronium ions in pure water. That would seem to make it more of a chemistry problem than a physics problem. Water dissociates slightly into
hydronium and hydroxide ions H3O+ and OH- at a concentration of
about 10E-7 moles per liter. Other than that?
 
GenMipps said:
I got the wrong answer somehow. Your hints were very helpful. Here is what I did:

(1 mol Water/18 grams Water) * (1 gram Water/1 cm3 Water) * (292 cm3 Water) * (6.022E23 molecules Water/1 mol Water) * (18 protons Water/1 molecule Water) * (1.602E-19 Coulombs/1 proton Water) giving me a result of 2.817E7 coloumbs, but the online grading thing says that's wrong. Did I mess something up?
The atomic weight of H2O is 18 but how many protons are there in a water molecule? (it is less than 18).

AM
 
The math is almost perfect except for the number of protons in water. Replace "18 protons Water" with 10 protons (Hyrdrogen= 1 proton, Oxygen=8; therefore, 10 total)
 
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