If 455-ml of 6.0 M HNO3 is used to make a 2.5 L dilution,what is the molarity of the dilution?
6.0 x 455/2500 =
Well, if we start with 455 ml of 6.0 M HNO3 and dilute it to 2.5 L, we can calculate the molarity of the dilution.
Let me just grab my calculator... *beep boop beep...*
Okay, so 455 ml is equivalent to 0.455 L. Now, we divide the amount of moles in the original solution by the final volume to get the molarity of the dilution.
And the answer is... *drumroll*... approximately 5.747 Molar (M).
So, the molarity of the dilution is about 5.747 M.
But hey, don't take my word for it. Double-check those calculations, just to be sure.
To find the molarity of the dilution, we need to use the formula:
M1V1 = M2V2
where:
M1 = initial molarity
V1 = initial volume
M2 = final molarity
V2 = final volume
Given that:
M1 = 6.0 M
V1 = 0.455 L
V2 = 2.5 L
Let's plug these values into the formula and solve for M2:
M1V1 = M2V2
(6.0 M)(0.455 L) = M2(2.5 L)
Now, let's solve for M2:
(6.0 M)(0.455 L) = M2(2.5 L)
2.73 mol = M2(2.5 L)
Divide both sides of the equation by 2.5 L to solve for M2:
(2.73 mol)/(2.5 L) = M2
M2 = 1.092 M
Therefore, the molarity of the dilution is approximately 1.092 M.
To find the molarity of the dilution, you can use the formula:
M1V1 = M2V2
where M1 and V1 are the initial molarity and volume of the solution, and M2 and V2 are the final molarity and volume of the dilution.
In this case, we can substitute the known values into the formula:
M1 = 6.0 M (the initial molarity of the HNO3 solution)
V1 = 455 ml (the initial volume of the HNO3 solution)
M2 = ? (the molarity of the dilution)
V2 = 2.5 L (the volume of the dilution)
First, we need to convert the initial volume to liters:
V1 = 455 ml = 0.455 L
Now we can plug these values into the formula and solve for M2:
(6.0 M)(0.455 L) = M2(2.5 L)
(6.0 M)(0.455 L) / (2.5 L) = M2
The calculation gives us:
M2 = 1.092 M
Therefore, the molarity of the dilution is approximately 1.092 M.