# Question 12530

Apr 10, 2017

The answer is c) $6.02 \cdot {10}^{24}$

#### 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 ${\text{18 g mol}}^{- 1}$, which implies that $1$ mole of water has a mass of $\text{18 g}$.

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 $6.02 \cdot {10}^{23}$ molecules of water $\to$ this is known as Avogadro's constant.

You can thus say that your sample contains $6.02 \cdot {10}^{23}$ 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

$\text{For H: } Z = 1$

$\text{For O: } Z = 8$

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

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

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)# ${\text{e}}^{-}$