Question

Question #78000

Answer 1

`"10. L"`

Explanation:

Start by calculating the number of moles of silver chloride present in the sample

`0.019 color(red)(cancel(color(black)("g"))) * "1 mole AgCl"/(143.32color(red)(cancel(color(black)("g")))) = 1.33 * 10^(-4)` `"moles AgCl"`

Now, silver chloride is an insoluble salt, which implies that when you dissolve silver chloride in water, most of the salt will remain undissociated.

An equilibrium will be established between the undissolved salt and the dissolved ion

`"AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-)`

The solubility product constant, `K_(sp)`, takes the form

`K_(sp) = ["Ag"^(+)] * ["Cl"^(-)]`

Notice that every mole of silver chloride that dissociates in aqueous solution produces `1` mole of silver(I) cations and `1` mole of chloride anions.

If you take `s` to be the molar solubility of the salt, i.e. the maximum concentration of silver chloride that dissolves in `"1 L"` of solution, you can say that you have

`K_(sp) = s * s`

`K_(sp)= s^2`

Rearrange to solve for `s`

`s = sqrt(K_(sp))`

Plug in your value to find

`s = sqrt(1.8 * 10^(-10)) = 1.34 * 10^(-5)`

This means that a saturated solution of silver chloride will contain `1.34 * 10^(-5)` moles of dissolved silver chloride for every `"1 L"` of solution, which, for all intended purposes, is equivalent to `"1 L"` of water.

Now, notice that your sample contains

`1.33 * 10^(-4)color(white)(.)"moles AgCl" ~~ color(red)(10) * overbrace((1.34 * 10^(-5)color(white)(.)"moles AgCl"))^(color(blue)("what can be dissolved in 1 L"))`

This implies that its volume will be `color(red)(10)` times that of a solution that contains `"1 L"` of water. You can thus say that your solution will contain

`color(darkgreen)(ul(color(black)("volume water = 10. L")))`

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