Answer:
[NH₄NO₃] at D → 0.279 M
Explanation:
This exercise involves a series of dilutions one after the other.
First of all, we calcualte ammonium nitrate's concentration at A.
15.71 g . 1 mol/ 80 g = 0.196 mol / 0.150 mL = 1.31 M
At B → 1.31 M . 20 mL/ 75 mL = 0.349 M
At C → 0.349 M . 15 mL / 25 mL = 0.209 M
[NH₄NO₃] at B = 0.349 M
[NH₄NO₃] at C = 0.209 M
So let's calculate the new moles
In 1 mL of B we have 0.349 mmoles
In 10 mL of B we have 3.49 mmoles
In 1 mL of C we have 0.209 mmoles
In 10 mL of C we have 2.09 mmoles
Volume of D = 10 ml + 10ml = 20 mL
Total mmmoles = 3.49 mmoles + 2.09 mmoles = 5.58 mmoles
[NH₄NO₃] at D = 5.58 mmoles / 20mL → 0.279 M
A . well test explanation for a broad set of observations.
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