Question

how many positive or negative charges are present in 300 gm if water ?

Answer 1

The question is: How many positive or negative charges are present in $300\mathrm{g}$ of water?

We have:

$$\vartheta (H _ {2} O) = \frac {3 0 0 g}{1 8 \frac {g}{m o l}} = 1 6. 6 7 m o l$$ $$1 m o l = 6. 0 2 * 1 0 ^ {2 3} m o l e c u l e s$$

So,

$$3 0 0 g = \frac {3 0 0}{1 8} m o l = \frac {3 0 0}{1 8} * 6. 0 2 * 1 0 ^ {2 3} m o l e c u l e s = \frac {3 0 0}{1 8} * 6. 0 2 * 1 0 ^ {2 3} * 1 0 e l e c t r o n s$$

(because 1 molecule of water = 1 + 1 + 8 electrons or protons)

Result:

$$3 0 0 g = 1. 0 0 3 * 1 0 ^ {2 6} \text {e l e c t r o n s (o r p r o t o n s)}$$