Answer:
0.182 g
Explanation:
0.436 g of the acid was dissolved in 60 mL of water and then transferred into a 100.00 mL beaker. The volume still remains 60 mL.
25 mL was then transferred to Erlenmeyer flask for titration.
If 60 mL of the solution contains 0.436 g of the unknown diprotic acid, then 25 mL of the same solution would contain:
   0.436 x 25/60 = 0.182 g
Hence, the mass of the diprotic acid in the 25.00 mL of the solution is 0.182 g.
Answer:
The 25 ml solution of an unknown diprotic acid contains 0.182 grams of a diprotic acid.
Explanation:
The diprotic acid dissolved in 60 ml water = 0.436 grams.
After the transfer of diprotic acid solution to the beaker, the volume of the solution remains 60 ml. So, the amount of diprotic acid = 0.436 grams.
On the transfer of 25 ml of the solution to the beaker, the amount of diprotic acid will be:
60 ml solution = 0.436 grams diprotic acid
25 ml solution = [tex]\frac{25}{60}\;\times\;0.436[/tex]
25 ml solution = 0.182 grams of diprotic acid.
The 25 ml solution of diprotic acid contains 0.182 grams of a diprotic acid.
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