Answer:
In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>
In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>
In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>
In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>
Explanation:
Using the dilution equation:
no of moles before dilution = no of moles after dilution.
Molarity x volume (initial)= Molarity x volume (final).
In order to prepare 10 mL, 5 μM from 25 μM solution,
Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?
25 x initial volume = 5 x 10
Initial volume = 50/25
= 2 mL
<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>
<em />
In order to prepare 10 mL, 10 μM from 25 μM stock,
25 x initial volume = 10 x 10
Initial volume = 100/25 = 4 mL
<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>
In order to prepare 10 mL, 15 μM from 25 μM stock,
25 x initial volume = 15 x 10
initial volume = 150/25 = 6 mL
<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>
In order to prepare 10 mL, 20 μM from 25 μM stock,
25 x initial volume = 20 x 10
initial volume = 200/25 = 8 mL
<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>
= 0.045 mol NH₃