Question

Question #b77a3

Answer 1

`1.4 * 10^(26)`

Explanation:

You know that your solution contains sodium chloride, `"NaCl"`, and water, so start by looking up the atomic numbers of the four elements present in the sample, i.e. sodium, chlorine, hydrogen, and water.

• `"Na: " Z = 11`
• `"Cl: " Z = 17`
• `"H: " Z = 1`
• `"O: " Z = 8`

As you know, the atomic number of an element tells you the number of protons located inside its nucleus. This means that your goal here will be to figure out how many atoms of sodium, chlorine, hydrogen, and oxygen are present in your solution.

Use the molar mass of water and the molar mass of sodium chloride to calculate the number of moles of water and of sodium chloride, respectively.

`58.5 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "1.001 moles NaCl"`

`360 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "19.98 moles H"_2"O"`

Next, use the chemical formulas of the two compounds to determine how many moles of each element are present in the sample.

Since `1` mole of sodium chloride contains `1` mole of sodium cations and `1` mole of chlorine anions, you can say that your sample contains `1.001` moles of sodium cations and `1.001` moles of chlorine anions.

Since you're interested in finding the number of protons present in the sample, you can treat the cations and the anions as atoms.

Similarly, `1` mole of water contains `2` moles of hydrogen and `1` mole of oxygen, you can say that your sample contains `19.98` moles of oxygen and `2 * 19.98 = 39.96` moles of hydrogen.

Next, use Avogadro's constant, `N_"A"`, to calculate the number of atoms of each element present in the sample.

`1.001 color(red)(cancel(color(black)("moles Na"))) * (N_"A"color(white)(.)"atoms Na")/(1color(red)(cancel(color(black)("mole Na")))) = (1.001 * N_"A")color(white)(.)"atoms Na"`

`1.001 color(red)(cancel(color(black)("moles Cl"))) * (N_"A"color(white)(.)"atoms Cl")/(1color(red)(cancel(color(black)("mole Cl")))) = (1.001 * N_"A")color(white)(.)"atoms Cl"`

`36.96 color(red)(cancel(color(black)("moles H"))) * (N_"A"color(white)(.)"atoms H")/(1color(red)(cancel(color(black)("mole H")))) = (36.96 * N_"A")color(white)(.)"atoms H"`

`19.98 color(red)(cancel(color(black)("moles O"))) * (N_"A"color(white)(.)"atoms O")/(1color(red)(cancel(color(black)("mole O")))) = (19.98 * N_"A")color(white)(.)"atoms O"`

Next, figure out the number of protons present for each element.

`(1.001 * N_"A") color(red)(cancel(color(black)("atoms Na"))) * "11 protons"/(1color(red)(cancel(color(black)("atom Na")))) = (11.011 * N_"A")color(white)(.)"protons"`

`(1.001 * N_"A") color(red)(cancel(color(black)("atoms Cl"))) * "17 protons"/(1color(red)(cancel(color(black)("atom Cl")))) = (17.017 * N_"A")color(white)(.)"protons"`

`(19.98 * N_"A") color(red)(cancel(color(black)("atoms O"))) * "8 protons"/(1color(red)(cancel(color(black)("atom O")))) = (159.84 * N_"A")color(white)(.)"protons"`

`(36.96 * N_"A") color(red)(cancel(color(black)("atoms H"))) * "1 proton"/(1color(red)(cancel(color(black)("atom H")))) = (36.96 * N_"A")color(white)(.)"protons"`

Next, add all the numbers to get the total number of protons present in your solution.

`[(11.011 + 17.017 + 159.84 + 36.96) * N_"A"]color(white)(.)"protons" = (224.83 * N_"A")color(white)(.)"protons"`

Finally, use the fact that

`N_"A" = 6.022 * 10^(23)`

to find

`"no. of protons" = 224.83 * 6.022 * 10^(23)color(white)(.)"protons"`

`color(darkgreen)(ul(color(black)("no. of protons" = 1.4 * 10^(26)color(white)(.)"protons")))`

The answer is rounded to two sig figs, the number of sig figs you have for the mass of water.