Question #41a9a

The thing to remember about a solution's normality is that it depends on the nature of the reaction.

In an acid-base reaction, a solution's normality is a measure of the concentration of hydronium cations, `"H"_3"O"^(+)`, or hydroxide anions, `"OH"^(-)`, present in the solution.

In your case, you know that the acid is dibasic, which implies that it can donate `2` protons in an acid-base reaction. This means that for every `1` mole of acid, you get `2` moles of hydronium cations.

So in this case, a `"0.1-M"` solution will have a normality of `"0.2 N"`.

Consequently, you can say that a `"0.1-N"` solution will have a molarity of `"0.05 M"`.

So all you have to do now is figure out how many grams of acid must be dissolved in `"100 mL"` of solution in order to get a solution that is `"0.05 M"`.

`100 color(red)(cancel(color(black)("mL solution"))) * overbrace("0.05 moles acid"/(10^3color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 0.05 M")) = "0.005 moles acid"`

To convert this to grams, use the molar mass of the acid

`0.005 color(red)(cancel(color(black)("moles acid"))) * "200 g"/(1color(red)(cancel(color(black)("mole acid")))) = color(darkgreen)(ul(color(black)("1 g")))`

The answer is rounded to one significant figure.