Dilution Lab Problems and Solutions (Answers)

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.

How to Solve Normality and Molarity of Solutions

Solving the Normality and Molarity of solutions is quite easy by remembering the relationship of Normality to Molarity.

1.    Normality may be equal but is always greater than the Molarity in the same solution.

2.    If the valence is 1, Normality is equal  to Molarity.

Try solving these problems. Atomic weights: Na = 23; Cl = 35.5, H = 1, Ca = 40, Valences: Ca = 2, NaCl = 1, HCl = 1.

1.    If you have dissolved 20 grams of sodium chloride in 1.5 Liter of distilled water, what is the:

1.1.    Normality
1.2.    Molarity
1.3.    Percent solution

2.    What is the Molarity of 1 N Hydrochloric acid?

3.    What is the Normality of 0.8 M calcium chloride?

CLICK HERE FOR THE ANSWERS.

Answers to Problem Solving, Normality, Molarity and Percent Solutions

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 %

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 %

HERE ARE THE ANSWERS

Problem Solving: Normality, Molarity, Percent Solution Problems

PROBLEM SOLVING : Normality, Molarity, Percentages

1.    WRITE DOWN FORMULA 1^ST.

2.    SHOW COMPUTATIONS.

3.    ENCLOSE ANSWERS IN BOXES.

ATOMIC WEIGHTS:

Na = 23           H = 1               S = 32

Cl= 35.5          O = 16            

1.       If you weigh 10 g NaCl solution and dissolve it in 600 ml of diluent. ( 5 pts. each)

A.    What is the Molarity?

M= 10

       58.5

      0.6 L

ANSWER= M = 0.28 M NaCl

B.     What is the Normality?

N= M X v or factor

N= 0.28 X 1

ANSWER N = 0.28 N NaCl

C.     What is the percent solution?

% = Weight of solute X 100

        Total Volume of solution

% = 10 X 100

          600

ANSWER % = 1.67%

2.      What is the Molarity of a 2 N H2SO4 solution? ( 5 pts.)

M= N

       v (valence or factor)

M= 2

       2

ANSWER M = 1 M H2SO4

3.      What is the dilution if you add 0.25 ml of serum to 5.0 ml of diluent? ( 5 pts.)

0.25 mL =  1 part

5.0 mL =  20 parts

               1:21

ANSWER Dilution = 1:21

Or

Parts of diluent = 5/0/25

5  

0.25

= 20 parts of diluent

DILUTION FACTOR= parts of solute + parts of diluent

Dilution factor = 1 + 20 = 21

ANSWER Dilution = 1:21

4.    How would you prepare  1:6 serum dilution? ( 5 pts.)

THERE ARE VARIOUS WAYS TO PREPARE SERUM DILUTIONS:

THE SIMPLEST WAY IS TO CONSIDER 1 mL of SERUM AS 1 PART.

ANSWERS:

1mL of serum = 1 part

5 mL of NSS  =  5 parts

                           6 parts

Dilution = 1:6

0.25 mL of serum   = 1 part

0.25 X 5 = 2.25 mL= 5 parts

                                   6 parts

Dilution = 1:6

5.    How would you prepare a 0.85 % NaCl solution? (5 pts.)

W=(%) X Total volume of solution

                100

W= 0.85 X 100

           100

ANSWER W= 0.85 g NaCl diluted to 100 mL of water in a volumetric flask

6. What is the Normality of a 10% CaCl2 solution?

N= % X 10

         EW (MW/v)

N= 10 X 10

        111/2

ANSWER N = 1.80 N CaCl2

7. Convert 40 mEq/L of calcium to mg/dL

40 mEq  X   1L    X  20 mg (40 atomic weight / valence)

     L            10 dL      1 mEq

ANSWER= 80 mg/dL

Laboratory Math: Percent Solutions, Normality, Molarity Problems, Answers To Practice Problems

ANSWERS TO PRACTICE PROBLEMS:

1.       IF YOU HAVE WEIGHED 90 GRAMS OF NAOH AND DISSOLVED IT 1,500 ML OF WATER, WHAT IS THE;

A.      PERCENTAGE

% = Weight in grams  X 100

       Total Volume in mL

% = 90 x 100

       1,500

% = 6 % of NaOH solution

B.      MOLARITY

M= W in grams

        MW

         Liter of solution

M = 90

        40

        1.5

M = 1.5 M NaOH

C.      NORMALITY

N= Mv

N = 1.5 X 1

N = 1.5 N NaOH

2.       HOW MANY GRAMS DO YOU NEED TO PREPARE 2M OF NACL?

W = DM X DV X MW

W = 2 X 1 X 58.5

W = 117 grams of NaCl

How: weigh 117 grams of NaCl and dilute it with DH20 up to the 1 Liter mark in a volumetric flask or measure 500 mL of DH2O into a volumetric flask, weigh 117 grams of NaCl and dissolve. Add DH20 up to 1 liter mark.

3.       HOW WOULD YOU PREPARE 1N OF HCL?

C1V1 = C2V2

C1= 12 N (Concentrated HCl)

C2 = 1N

V2 = 100 mL (assumption when no volume is given)

V1 = unknown

V1 = C2V2

           C1

V1 = 8.33 mL

How: Add 8.33 mL of concentrated HCl to 91.67 mL of DH2O.

4.       HOW WOULD YOU PREPARE  0.85% NACL SOLUTION?

W = % X TV (total volume)

         100

W = 0.85 X 100 (if no volumes are given assume that TV is 100)

            100

W = 0.85 g (grams) of NaCl

How: Measure 50 mL of DH2O into a volumetric flask. Weigh 0.85 grams of NaCl and dissolve completely. Add DH2O up to the 100 mL mark.

5.       WHAT IS THE NORMALITY OF A 0.5 M CACL2 SOLUTION?

N = M X v (factor)

N = 0.5 X 2

N = 1 N CaCl2

6.       HOW DO YOU PREPARE A 1:10 SERUM DILUTION?

When no volume is given assume that 1 mL is one part; hence,

1 mL = 1 part of solute

9 mL = 9 parts of diluent

           10 parts (total)

Hence: dilution is 1:10

How: Add 1 mL of serum to 9 mL of NSS to come up with a 1:10 serum dilution.

7.       MAKE UP 500 ML OF 5M SOLUTION OF HCL

C1V1 = C2V2

V1 = C2V2

           C1

Before you substitute be sure that all your units are the same. It does not matter what units you used as long as they are similar units. i.e. if you use liters for volume all volumes should be expressed in liters.

V1 = 5 M X 500 mL

             12 (12 is standard Normal concentration of concentrated HCl, since valence is 1, N=M.)

V1 = 208.33 mL

How: Add 208.33 mL to 291.67 of DH2O to produce 500 mL of 5M HCl solution.

8.       PREPARE THE FOLLOWING WORKING STANDARD  SOLUTIONS FROM A STOCK SOLUTION OF 20 MG/DL: STATE VOLUME OF DILUENT AND DILUTION.

A.      15 MG/DL

                 C1V1 = C2V2

                 V1 = C2V2

                            C1

                V1 = 15 mg/Dl X 100 mL

                            20 mg/dL

                V1 = 75 mL 

75 mL of 20 mg/dL stock solution added to 25 mL of DH2O

B.      10 MG/DL

               C1V1 = C2V2

               V1 = C2V2

                          C1

                V1 = 10 mg/Dl X 100 mL

                            20 mg/dL

                V1 = 50 mL of 20 mg/dL stock solution added to 50 mL of DH2O

C.      5 MG/DL

              C1V1 = C2V2

               V1 = C2V2

                          C1

                V1 = 5 mg/Dl X 100 mL

                            20 mg/dL

                V1 = 25 mL of 20 mg/dL stock solution added to 75 mL of DH2O

D.      2 MG/DL

C1V1 = C2V2

V1 = C2V2

   C1

V1 = 2 mg/Dl X 100 mL

                            20 mg/dL

V1 = 10 mL of 20 mg/dL stock solution added to 90 mL of DH2O

Volume of stock solution, diluent and dilution are the following:

Volume of stock solution	Volume of Diluent (DH2O)	Dilution
75 mL	25 mL	1:4 (diluent to solute)
50 mL	50 mL	1:2
25 mL	75 mL	1:4
10 mL	90 mL	1:9

1.       75 divided by 25 = 3 + 1= 4, hence dilution is 1:4 (diluent to solute)

2.       50 divided by 50 = 1 + 1= 2, hence dilution is 1:2

3.       75 divided by 25 = 3 + 1= 4 hence dilution is 1:4

4.       90 divided by 10 = 8 + 1 = 9, hence dilution is 1:9