# Charge in water

1. Aug 28, 2006

### Valhalla

The problem states: Calculate the number of coulombs of positive charge in 250 cm3 of (neutral) water. (Hint: A hydrogen atom contains one proton and an oxygen atom contains eight protons.)

Ok, first I converted to moles. Using the density of water.

$$250mL * \frac{1g}{1mL} * \frac{1 mole H_2O}{18.02g}=13.8735moles H_2O$$

$$13.8735molesH_2O*6.022*10^{23}=8.3546*10^{24} atoms H_2O$$

$$8.3546*10^{24}*(2*(1 proton H)+8 proton O)=8.3546*10^{25} protons$$

$$8.3546*10^{25} protons * 1.6*10^{-19} Coulombs=1336736Coulombs$$

Does this seem right? That seems really large. However, the electrons calculate up to exactly the same amount and cancel it out.

Last edited: Aug 28, 2006
2. Aug 28, 2006

### Bystander

3. Aug 28, 2006

### Valhalla

oh ok, I didn't check the second page. Thanks!