Problem Description:
You are a workstudy for the chemistry department. Your supervisor has just asked you to prepare 500mL of 3M HCl for tomorrow's undergraduate experiment. In the stockroom explorer, you will find a cabinet called "Stock Solutions". Open this cabinet to find a 2.0L bottle labeled "11.6M HCl". The concentration of the HCl is 11.6M. Please prepare a flask containing 500 ml of a 3 M (+/- 0.005M) solution and relabel it with its precise molarity. Note that you must use realistic transfer mode, a buret, and a volumetric flask for this problem.
As a reminder, to calculate the volume needed to make a solution of a given molarity, you may use the following formula:
C1V1 = C2V2
1. Calculations: (3 marks)
C1V1 = C2V2
(11.6M)(V1) = (3M)(500ml)
11.6M V1 = 1500(M x ml)
V1 = 129.31ml
Volume of HCl used: 129.31mL
2. How do you calculate your percent error?
% error = (experimental concentration – 3.0 M) / 3.0 M * 100, in absolute value.
You need1.5 mols of HCl
You have 11.6 mols per liter
1.5 /11.6 = .1293 liters of the strong stuff (agree with you)
add that to .5 - .1293 = .3707 liters of neutral water
and you have 0.5 Liters = 500 mL of 3M HCl
now you need to know that strong bottle with the 11.6 M percent error to get the error in your result I think but anyway what is your measurement error for the .1293 liters?
say for example it was .01 liters (10 mL)
then you would end up with .1393 liters of strong which would be 11.6 liters*.1393 mols/liter = 1.616 mols instead of 1.5 mols
that is 100 (1.616 - 1.5) /1.5 = 7.7 % error
I don't see a question.
I don't see anything wrong with what you've done except you didn't explin how to prepare the solution.
You want to add the 129.31 mL of 11.6 molar solution to the 500 mL volumetric flask, add DI water to the mark on the flask, mix, stopper, label.
What you have labeled as absolute error actually is percent error since you have multiplied by 100.
To calculate the percent error for a solution preparation, you need to compare the experimental concentration with the desired concentration and calculate the difference as a percentage.
The formula to calculate percent error is:
% error = (experimental concentration - desired concentration) / desired concentration * 100
In this case, the desired concentration is 3.0 M. The experimental concentration can be calculated as follows:
Experimental concentration = total moles of solute / total volume of solution
Since the experimental concentration is not directly given in the problem, we need to first calculate the moles of solute used and then divide it by the total volume of the solution.
To find the moles of solute used, we can use the equation C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
Using the given values:
Initial concentration (C1) = 11.6 M
Initial volume (V1) = 129.31 mL (calculated from the previous step)
Final concentration (C2) = 3.0 M
Final volume (V2) = 500 mL
Now, we can calculate the moles of solute used:
Moles of solute = C1V1 = (11.6 M)(129.31 mL) = 1500 M·mL
Convert the volume to liters for mole calculation:
1500 M·mL = 1.5 M·L
Now, we have the moles of solute used. Next, we need to calculate the experimental concentration by dividing the moles of solute by the total volume of the solution:
Experimental concentration = Moles of solute / Total volume of solution
Experimental concentration = 1.5 M·L / 0.5 L (since the total volume should be in liters)
Calculating the experimental concentration:
Experimental concentration = 1.5 M·L / 0.5 L = 3.0 M
Now that we have the experimental concentration, we can calculate the percent error using the formula mentioned earlier:
% error = (experimental concentration - 3.0 M) / 3.0 M * 100 (in absolute value)
% error = |(3.0 M - 3.0 M) / 3.0 M| * 100
% error = |0 / 3.0 M| * 100
% error = 0%
Therefore, the percent error is 0%, indicating that the experimental concentration matches the desired concentration exactly.