<u>According to the Question:</u>
<u>We initially had 0.448 grams of the unknown diprotic acid, which was added to about 60 mL of water and that solution was added to a 100 mL flask and filled to the 100 mL mark</u>
The above is as mentioned in the question, our 0.448 grams of diprotic acid is basically diluted to 100 mL <em>[it's volume has been increased to 100mL]</em>
but the amount of the acid is still the same
Which means that we now have 0.448 grams of the diprotic acid in a 100mL solution
<u>Percent of the diprotic acid that will be present in the Erlenmeyer Flask:</u>
From this 100 mL solution, 25 mL is transferred to the Erlenmeyer Flask
Which means that<em> (25/100) * 100 =</em> 25 % of the 0.448 grams of diprotic acid is present in the 25 mL sample
<u>Mass of di-protic acid in the 25 mL solution:</u>
Since 25% of the initial amount remains in the final solution,
Mass of the diprotic acid in the final solution = 0.448 * 0.25
Mass in the final solution = 0.112 grams
Therefore, 0.112 grams of the di-protic acid will be present in the Erlenmeyer Flask
