Dilution Lab Problems and Solutions (Answers)
In the laboratory, you are often asked to prepare dilutions of solutions. These dilution lab problems can sometimes leave you grasping for solutions. Well, fret no more, in this post, simple dilution problems are given with the corresponding answers below. I hope you enjoy this learning process.
Instructions:
For the following problems, identify the given and the unknown. State what formula could be used, and show your computations.
1. How do you prepare a 1:4 dilution of HCl?
2. What’s the volume of diluent (NSS) needed to prepare a 1:5 serum dilution with a total of 5 mL?
3. How do you prepare a Normal Saline Solution (NSS)?
4. What is the resulting dilution for each tube in this serial dilution?
Tube No. Volume of stock standard solution in mL Volume of diluent in mL Resulting Dilution
1
2
3
5. How much volume do you need to prepare a 5 ml of a 1:10 dilution?
ANSWERS
1. How do you prepare a 1:4 dilution of HCl?
A 1:4 dilution indicates that there for every 1 part of the solute, there are 3 parts of the solvent. 4 indicate the total of the parts of the solute and the solvent. The solute is HCl (hydrochloric acid) and the solvent would be distilled water.
The easiest method is to assume that I part = 1 mL (milliliter), hence, 1 part is = 1mL HCl
If 1 part = 1 mL
Hence:
3 parts = 3 mL
When you add the parts, the total is 4, hence, the dilution 1:4
So, to prepare 1:4 dilution, you add 1 mL of HCL to 3 mL of distilled water.
2. What’s the volume of diluent (NSS) needed to prepare a 1:5 serum dilution with a total of 5 mL?
NSS or Normal Saline Solution is also 85% saline. This is also very simple to solve, since 5 is the total of the solute and diluent parts, you can assume that 1 part = 1 mL.
So, 4 parts is required to complete the 5 parts.
Since 1 part =1 mL
So, 4 parts = 4 mL
Since the total volume is also 5 mL, then by adding 1 mL of serum + 4 mL of NSS; you, therefore, need 4 mL of NSS to prepare the dilution.
3. How do you prepare a serum dilution of 1:3, if your available serum is only 0.25 mL?
1:3 – indicates 1 part of serum + 2 parts of diluent
Since the available solute or serum volume is only 0.25 mL, you have to equate this to 1 part of the dilution.
Hence, if 1 part = 0.25 mL
Therefore, 2 parts diluent = 0.25 mL x 2 parts = 0.50 mL
Hence, to prepare a serum dilution of 1:3, you can add:
0.25 mL of serum + 0.50 mL of diluent.
1 part + 2 parts = 3 parts
So, the dilution is 1:3
4. What is the resulting dilution for each tube in this serial dilution?
Tube No. Volume of standard solution in mL Volume of diluent in mL Resulting Dilution
1 0.5 (pure stock soln.) 1 1:3
2 0.5 from tube #1 1 1:9
3 0.5 from tube #2 (mix and discard 0.5 mL) 1 1:27
5. How much volume do you need to prepare a 5 ml of a 1:10 dilution of standard solution?
1:10 dilution indicates what?
Yes, it indicates that 1 part of standard stock solution is added to 9 parts of diluent.
You can do the easiest method by adding 1 mL of the standard stock solution plus 9 mL of the diluent.
However, since the total volume is stated, which is 5 mL, you have to determine how many parts would the solution consists of.
You can divide 5 mL by 10, to determine the volume of each part.
Hence, 5/10 = 0.5 mL
So, you can now equate 1 part with 0.5 mL
The 9 parts diluent would therefore be = 9 x 0.5 = 4.5 mL
So, you add 0.5 mL standard stock solution to 4.5 mL diluent to come up with a 5 mL total volume of 1:10 standard dilution.
There you go! It’s relatively easy, if you always remember that the dilution factor (DF) is the total of 1 part of the solute and the designated parts of the solvent.
If I say the DF is 5, I’m also stating that the dilution is 1:5. So, there are 1 part of solute + 4 parts of solvent.
For 1:9
There is one part of solute + 8 parts of the solvent
Dilution is different from ratio because in ratio the numbers remain the same. You don’t add them. Unlike in dilution, you add the parts of the solute plus parts of the solvent.
If you’re performing serial dilution, remember to multiply the previous dilution of the tube from where you got the solute.
Good luck with your laboratory math during your exams or when you’re working.
Showing posts with label clinical chemistry problem solving. Show all posts
Showing posts with label clinical chemistry problem solving. Show all posts
Sunday, December 24, 2017
Thursday, September 25, 2014
Practice Problems: Molarity, Normality and Percent Solutions; Laboratory Math
1. What is the Normality of a 3.6 M Sulfuric acid solution given the
molecular weights?( H=1, S=32, O-16)
a. 1.8 N
b. 3.6N c. 4.9 N d. 7.2 N e.
NIL
2. How many grams of HCL is used to
prepare 250 ml of a 4.8 solution of HCL? (H-1, Cl-35.5)
a. 36
G b. 36.5 g c.
40 g d. 43.8 g e. NIL
3. What is the Molarity of a 2 .5 N solution
of NaOH?
a.
1.50 mol/l c. 3.35 mol/l e.
NIL
b.
2.5 mol/l d. 4.5 mol/l
4. The molecular weight of H3 PO4 is:
a. 48 c.
98 e. NIL
b. 58.5 d.
98.1
5. How many grams
of NaCl are required to make 1,000 ml.
of 0.3 M
solution?
a. 36.3 c.
26.32 e. NIL
b. 52.6 d. 53
6. What is the
Molar concentration of a 20 grm. Of NaOH diluted to 1 liter of distilled water?
a. 1 M c.
0.5 M e. NIL
b. 2 M d.
1.5 M
7. What is the amount of CaCl2.H2O in grams is
needed to prepare 0.4 N solution of
Ca Cl2?
a. 8.8 c. 16 e. NIL
b. 4 d.
32
8. One milligram is equal to:
a. 0.001 grm. c.
0.01 grm. e. NIL
b. 0.0001 grm. d.
0.1 grm.
9. A 10 mgs. % solution contains:
a. 10 mgs of solute/100ml of solution
b.
10 mgs
of solute/100 ml of diluent
c.
10 mgs
of solute/1000 ml of diluent
d.
10 mgs
of solute/1000ml of solution
e.
NIL
10. A 2 % solution of 10mg/100ml is diluted
1:100. What is the final concentration?
a. 2 % c.
0.2 % e. NIL
b. 0.02 % d.
0.002 %
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