Question

Question #1f4c5

Answer 1

`"4.00 mol L"^(-1)`

Explanation:

For starters, you know that the density of pure water is approximately equal to `"1.00 g mL"^(-1)`. By comparison, this aqueous solution has a density of `"1.25 g mL"^(-1)`.

The idea here is that the difference between the mass of `"1 mL"` of pure water and the mass of `"1 mL"` of this solution will represent the mass of the solute.

For `"1 mL"` of this solution, you have

`overbrace("1.25 g")^(color(blue)("mass of solution")) - overbrace("1.00 g")^(color(blue)("mass of solvent")) = overbrace("0.25 g")^(color(blue)("mass of solvent"))`

Use the molar mass of the solvent to find the number of moles present in `"1 mL"` of this solution.

`0.25 color(red)(cancel(color(black)("g"))) * "1 mole solute"/(62.5color(red)(cancel(color(black)("g")))) = "0.00400 moles solute"`

As you know, the molarity of the solution tells you the number of moles of solute present in `10^3color(white)(.)"mL" = "1 L"` of the solution.

Since you know that `"1 mL"` contains `0.00400` moles of solute, you can say that `10^3` `"mL"` will contain

`10^3 color(red)(cancel(color(black)("mL solution"))) * "0.00400 moles solute"/(1color(red)(cancel(color(black)("mL solution")))) = "4.00 moles solute"`

You can thus say that the molarity of the solution is equal to

`color(darkgreen)(ul(color(black)("molarity = 4.00 mol L"^(-1))))`

The answer is rounded to three sig figs.