What volume of 3.16 M HNO3 is required to
prepare 524 mL of a 0.2 M HNO3 solution?
Answer in units of mL.
524 mL = 0.524 L
mols you need = M x L = 0.2 x 0.524 = 0.104
Then M = mols/L or
3.16M = 0.104/L so L = 0.104/3.16 = 0.03316 L or 33.16 mL.
We're allowed only three significant figures; therefore, I would round that to 33.2 mL.
You may find it easier to use the dilution formula.
c1v1 = c2v2
0.2 x 524 = 3.16 x v2
v2 = (0.2*524/3.16) = 33.2 mL.
How many moles do you want? That's M x L = ?
Then M HNO3 = mols HNO3/L HNO3. You know M and mols. Solve for L and convert to mL.
Not even close. Post your work and I'll find your error.
I converted my 524ml to liters so that became 524000. Then I did 3.16x1655.84. Then converted back to liters. But i just realized i skipped a step, but I don't understand where I get the numbers at for that.
thanks cheers mate that was helpful for me good job
To calculate the volume of a solution needed, you can use the formula:
(V1)(C1) = (V2)(C2)
Where:
V1 is the volume of the initial solution
C1 is the concentration of the initial solution
V2 is the volume of the final solution
C2 is the concentration of the final solution
In this case, we are looking for V1, the volume of the 3.16 M nitric acid (HNO3) solution. We know:
V2 = 524 mL (the final volume of the solution)
C2 = 0.2 M (the final concentration of the solution)
C1 = 3.16 M (the initial concentration of the solution)
Now we can substitute these values and solve for V1:
(V1)(3.16 M) = (524 mL)(0.2 M)
Simplifying the equation:
V1 = (524 mL)(0.2 M) / (3.16 M)
Calculating the volume:
V1 = 33.1 mL
Therefore, you will need 33.1 mL of the 3.16 M HNO3 solution to prepare 524 mL of a 0.2 M HNO3 solution.