0.083 L of concentrated 12.0M HNO3
Explanation:This is a dilution chemistry problem. For problems like this one, use the following equation
aaaaaaaaaaaaaaaaM1V1=M2V2
Where
M1=molarity of solution 1
V1=volume of solution 1
M2=molarity of solution 2
V2=volume of solution 2
Whenever you dilute a concentrated stock solution, or for that matter, any solution, you are essentially changing the volume of that solution by adding water. If we look at what Molarity exactly is,
Molarity=moles1 Liter solution
you can see when you increase the volume, you decrease the molarity(concentration). In dilution cases, you are adding water so moles of the solute do not change, only the volume of the solution changes.
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Using our equation from above, we can find the number of moles of nitric acid, HNO3, from what we are given.
We will let M1 and V1 be 8.0M and 125 mL respectively. Make sure to convert mLto L first. For M2, use the 12.0M. Isolate V2 and solve.
M1V1=M2V2(0.125L)(8.0M)=(12.0M)(V2)(0.125L)(8.0M)12.0M=V20.083 L=V2Answer: 0.083 L
