How many elections are present in 90 gram of water
Answers
Answer:
Let’s say it is pure water with a mass m = 90 g. First we want to know the number N of water molecules. For that we use the molar mass of water which is close to M = 18 g/mol. Therefore there is m/M = 5 moles of water molecules. One mole is 6.022 x 10^23 (the Avogadro number). This leads to N = 5 x 6.022 x 10^23 ~ 3 x 10^24 molecules.
For each water molecules (H2O), there is 10 electrons: 8 for the oxygen and one for each hydrogen (there are sharing the electrons but at the end it is neutral so we can count the protons and not care about the details of the chemistry).
At the end, in 90 g of pure water, we have something close to 5 x 10^25 electrons. If there is some ions in the water (which is likely), it would change a little bit this number (in this calculus, at some point we multiplied by the number of protons per molecules and we divided by the number of protons + neutrons. This ratio depends a bit on the atoms that you consider) but not much so the order of magnitude is correct anyway.
Explanation: