Question

Question #12530

Answer 1

The answer is c) `6.02 * 10^(24)`

Explanation:

The idea here is that you must use the density of water to convert the given volume to grams

`18 color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = "18 g"`

Now, the molar mass of water is approximately equal to `"18 g mol"^(-1)`, which implies that `1` mole of water has a mass of `"18 g"`.

You can thus say that your sample of water is equivalent to `1` mole of water. As you know, `1` mole of water must contain `6.02 * 10^(23)` molecules of water `->` this is known as Avogadro's constant.

You can thus say that your sample contains `6.02 * 10^(23)` molecules of water.

To find the number of electrons present in the sample, calculate the number of electrons present in `1` molecule of water. To do that, grab a Periodic Table and look for the 8atomic numbers* of the two elements that make up the water molecule

`"For H: " Z = 1`

`"For O: " Z = 8`

As you know, a neutral atom has equal numbers of protons inside its nucleus and electrons surrounding its nucleus. This implies that `1` molecule of water will have a total of

`"no of e"^(-)color(white)(.)"in 1 molecule" = overbrace(2 xx "1 e"^(-))^(color(blue)("from 2 atoms of H")) + overbrace(1 xx "8 e"^(-))^(color(purple)("from 1 atom of O")) = "10 e"^(-)`

Therefore, the total number of electrons present in your sample will be equal to

`6.02 * 10^(23) color(red)(cancel(color(black)("molecules H"_2"O"))) * "10 e"^(-)/(1color(red)(cancel(color(black)("molecule H"_2"O")))) = 6.02 * 10^(24)` `"e"^(-)`