Question

Will someone please help me with this? Thank you so much!!!

Answer 1

Well, let's go back through `1 - 5` then. You'll have to be careful with sig figs... I've given myself one more sig fig because `"0.1 M"` is just sad. I've seen two sig figs very often on concentrations.

I got `4.9%`.

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The mols of `"NaOH"` are found simply by the volume of it added to acetic acid.

`"0.10 mol NaOH"/cancel"L" xx 74.1 cancel"mL" xx cancel"1 L"/(1000 cancel"mL") = ul"0.0074(1) mols"`

And they react in a `1:1` ratio, so this is also the mols of acetic acid.

Therefore, in the `"10 mL"` of vinegar, `"0.0074(1) mols"` of it was acetic acid that reacted, resulting in a molarity for trial 1 of

`"0.0074(1) mols acetic acid"/"0.010 L" = ul"0.74(1) M"`

and in trial 2 a molarity of

`"0.0090(2) mols acetic acid"/"0.010 L" = ul"0.90(2) M"`

for an average molarity of

`("0.74(1) M" + "0.90(2) M")/2 = ul"0.82(2) M"`

So, the grams of acetic acid per liter is just the mass of it in `"1 L"`:

`(0.82(2) cancel"mols acetic acid")/"L" xx ("60.05 g CH"_3"COOH")/cancel("1 mol") = ul"49.(3) g/L"`

And so, the mass percent in vinegar is the mass of acetic acid in `"1 L"` divided by the mass of `"1 L"` of vinegar.

The mass of `bb"1 L"` of vinegar is:

`cancel"1 L" xx "1.005 g"/cancel"mL" xx (1000 cancel"mL")/cancel"L" = ul"1005 g vinegar"`

And so, the mass percent is:

`color(blue)(%"w/w") = "49.(3) g acetic acid"/"1005 g vinegar" xx 100%`

`= color(blue)ul(4.9%)`