500 mL is half a liter
so
We need 1.5 mols of HCl
we have a bottle containing 11.6 mols in every liter
so we need 1.5 /11.6 liters from that strong stuff bottle
That is pour 0.1293 liters (that is 129 mL) from that strong stuff bottle into an empty bottle. (agree with you)
now you have a bottle with 11.6 mols of HCl in it. fill it up to the 500 mL line with clean water. That means you add 500 -129 = 371 mL of clean water.
fill it up to the 500 mL line
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
Is this how I would do this problem? If not please explain how, thanks!
9 answers
So volume of HCl is 129.3mL?
The information here is inconsistent because we want a final concn of HCl to be 3 M with +/-0.005 M accuracy. So I would want to know that the 11.6 M stock solution was 11.60 M and I would want to end up with 3.000 M +/- 0.005. All of this has a significant figure problem.
Personally, I think this exercise is designed for you to describe how to make the solution. Add129.3 mL of the 11.6 (measure the 129.3 mL with a buret), place that in a 500 mL volumetric flask, add ENOUGH WATER TO REACH THE LINE ON THE VOLUMETRIC FLASK. That may or may not be 371 mL of H2O and in fact you don't care how much water it is because the TOTAL needs to be 500. Technically you don't know that 129 mL of the stock plus 371 mL H2O will equal 500. (This last point, I believe, is the major point of the problem. Ignoring the significant figure problem, you can calculate the final concn as
(11.6 M x 129.3)/500 = 2.999 which is within the +/- 0.005. Another point that should be made is that 500 mL volumetric flask are not exactly 500 mL.(buy close--within 0.5 mL).
I suppose the bottom line is that I have trouble trying to make a solution to within +/- 0.005 M when the directions don't contain values with that accuracy.
Personally, I think this exercise is designed for you to describe how to make the solution. Add129.3 mL of the 11.6 (measure the 129.3 mL with a buret), place that in a 500 mL volumetric flask, add ENOUGH WATER TO REACH THE LINE ON THE VOLUMETRIC FLASK. That may or may not be 371 mL of H2O and in fact you don't care how much water it is because the TOTAL needs to be 500. Technically you don't know that 129 mL of the stock plus 371 mL H2O will equal 500. (This last point, I believe, is the major point of the problem. Ignoring the significant figure problem, you can calculate the final concn as
(11.6 M x 129.3)/500 = 2.999 which is within the +/- 0.005. Another point that should be made is that 500 mL volumetric flask are not exactly 500 mL.(buy close--within 0.5 mL).
I suppose the bottom line is that I have trouble trying to make a solution to within +/- 0.005 M when the directions don't contain values with that accuracy.
what is the percentage error?
% error = (experimental concentration – 3.0 M) / 3.0 M * 100, in absolute value.
% error = (experimental concentration – 3.0 M) / 3.0 M * 100, in absolute value.
8==3
ngh i cant find the "Calculate your percent error" anywhere
;-;
;-;
% error = (experimental concentration – 3.0 M) / 3.0 M * 100, in absolute value. anyone know the percent error
Also some people are saying that 500 ml is the hcl used is that the answer
may i know the proper procedure to do this experiment?
2 answers