A) How do you make 200 mL of a 0.1 M solution of a substance that has a molecular weight of 121.1 g/mol?
Answer I got: 200mL = 1/0.1M x 1/(121.1g/mol) x mass of the substance.
mass of the substance = 0.2L x 0.1M x 121.1g/mol=2.42g
Note that my answer doesn't reveal how to do it?
B) If you take 10 mL of the solution you made in A, add 90 mL of water, mix and then take 5 mL of the mixture and bring it to 25 mL, what will be the concentration of the final solution in molars, millimolars, and micromolars?
Answer I used the multiple dilution equation first:
D1: 100/10= 10
D2: 25/5= 5
Df total: 5X10= 50 to 1
then I did this:
C2= C1/Df total:
C2= 0.1/50= .002 M;
.002/10^-3= 2 mM;
2/10^-6= 2 x 10^6 micromoles
This is not correct because the second dilution factor should be (30mL/5mL)=6mL. The dilution factor is the total volume divided by the initial added. Therefore the final answer is .0017M, 1.7 millimolar, and 1.7x10^3 micromolar.
Please be sure to post the actual subject in the subject line so that people don't waste time looking at posts out of their areas.
Thanks.
#1 has the correct number of grams. Having noted that the answer doesn't reveal how to do it, what do I do about it? Do you know how to make it up or do you want me to tell you how to make it up?
#2. M is ok at 0.002 M
That is 2 mM.
But it is 0.002 M x (1 M/10^6 micromols) = 2 x 10^-3?
Check my work.
Would you inform me on how this is made up. I didn't know how to make it up once I found the grams or if I was even supposed to find grams. Secondly, isn't the last answer 2 x 10^3 because a micromole is 10^-6 so it would be .002M/10^-6= 2x10^3. Right?
Thanks
You are correct on the micromolar. Note I didn't do the conversion properly. The units don't turn out right, either.
0.002 M x (1 x 10^6 micromoles/1 M) = 2000 micromoles.
How do we make the 0.1 M solution?
Take 2.42 g of the material, dissolve it in a little water, then make the final volume to 200 mL. Note that this is NOT the same thing as adding 2.42 g of the material to 200 mL water. In the first case, the final volume is 200 mL. In the second case, the final volume is more than 200 mL. In making molar solutions, one ALWAYS dissolves the substance in water, THEN make to some predetermined final volume.
A) To make a 200 mL of a 0.1 M solution of a substance with a molecular weight of 121.1 g/mol, you need to follow the formula:
Mass of the substance = Volume x Concentration x Molecular weight
In this case, the volume is 200 mL, the concentration is 0.1 M, and the molecular weight is 121.1 g/mol. Plugging in these values, you get:
Mass of the substance = 0.2 L x 0.1 M x 121.1 g/mol = 2.42 g
So, you need to measure out 2.42 grams of the substance and dissolve it in enough solvent (usually water) to make a total volume of 200 mL.
B) To find the concentration of the final solution after diluting 10 mL of the solution made in part A with 90 mL of water, and then bringing 5 mL of the resulting mixture to a total volume of 25 mL, you can use the dilution equation:
CiVi = CfVf
where Ci and Vi are the initial concentration and volume, and Cf and Vf are the final concentration and volume.
For the first dilution (10 mL solution + 90 mL water):
Ci = 0.1 M
Vi = 10 mL
Cf = unknown
Vf = 100 mL (10 mL + 90 mL)
Using the dilution equation:
Cf x Vf = Ci x Vi
Cf x 100 mL = 0.1 M x 10 mL
Cf = (0.1 M x 10 mL) / 100 mL
Cf = 0.01 M
For the second dilution (5 mL mixture + 20 mL water):
Ci = 0.01 M
Vi = 5 mL
Cf = unknown
Vf = 25 mL (5 mL + 20 mL)
Using the dilution equation again:
Cf x Vf = Ci x Vi
Cf x 25 mL = 0.01 M x 5 mL
Cf = (0.01 M x 5 mL) / 25 mL
Cf = 0.002 M
Converting to millimolars and micromolars:
1 mM = 0.001 M
1 μM = 0.000001 M
0.002 M = 2 mM
0.002 M = 2 x 10^6 μM
Therefore, the concentration of the final solution is 2 millimolars (2 mM) and 2 million micromolars (2 x 10^6 μM).