If you dilute 14.0mL of the stock solution to a final volume of 0.350L , what will be the concentration of the diluted solution? Before it, there was a question: How many milliliters of a stock solution of 5.60 M HNO3 would you have to use to prepare 0.190L of 0.450 M HNO3? and the answer is 15.3 mL.
Well, it looks like you're having quite the chemistry adventure! So, you've successfully prepared a 0.450 M HNO3 solution using 15.3 mL of a stock solution. That's impressive! Now, onto the next challenge.
To calculate the concentration of the diluted solution, we need to use the dilution equation, which is like a chemistry math puzzle. So, grab your lab coat and let's solve this!
We know that the initial volume is 14.0 mL, and the final volume is 0.350 L. Now, all we need to do is divide the initial concentration by the final volume to find the concentration of the diluted solution.
Let's plug in the numbers: 14.0 mL / 0.350 L = 0.040 M.
So, the concentration of the diluted solution is 0.040 M. Ta-da! We've solved the concentration riddle. Keep up the fantastic work, chemistry explorer!
To find the concentration of the diluted solution, you can use the formula:
initial concentration × initial volume = final concentration × final volume
For the first part of your question, the initial concentration is 5.60 M (as given), the initial volume is 15.3 mL (as calculated from the previous question), and the final volume is 0.190 L:
(5.60 M) × (15.3 mL) = (final concentration) × (0.190 L)
Now, let's rearrange the equation to solve for the final concentration:
(final concentration) = (5.60 M × 15.3 mL) / (0.190 L)
Calculating this expression:
(final concentration) = 90.48 mL·M / 0.190 L
(final concentration) = 475.7 M
Therefore, the concentration of the diluted solution is 475.7 M.
To find the concentration of the diluted solution, you need to use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the stock solution
V1 = volume of the stock solution used
C2 = final concentration of the diluted solution
V2 = final volume of the diluted solution
In this case, the initial concentration of the stock solution (C1) is 5.60 M, and the final concentration of the diluted solution (C2) is unknown. The volume of the stock solution used (V1) is 15.3 mL, and the final volume of the diluted solution (V2) is 0.190 L.
Plugging these values into the formula:
(5.60 M)(15.3 mL) = C2(0.190 L)
First, convert the volume from milliliters (mL) to liters (L):
(5.60 M)(0.0153 L) = C2(0.190 L)
Next, rearrange the formula to solve for C2:
C2 = (5.60 M)(0.0153 L) / (0.190 L)
C2 ≈ 0.454 M
Therefore, the concentration of the diluted solution is approximately 0.454 M.
if the stock solution is 0.450M, then if you add x mL of water, you have
0.450(14.0) = 0.350(14.0+x)
x = 4.0
check: the concentration is to be reduced by a factor of .35/.45, so the volume must be increased by a factor of .45/.35
.45/.35 * 14 = 18, so we added 4 mL of water.