Question #12530

1 Answer
Apr 10, 2017

Answer:

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"^(-)#