ANSWERS TO NORMALITY, MOLARITY AND PERCENT SOLUTION QUESTIONS

2. What is the Molarity of a 0.5 N H2SO4 solution? (5 pts.)
3. What is the dilution if you add 0.75 ml of serum to 3.0 ml of diluent? ( 5 pts.)
4. How would you prepare 1:20 serum dilution? (5 pts.)
5. How would you prepare a 0.5 % NaCl solution? (5 pts.)
6. You are asked to prepare 200 ml of the following standard concentrations from a 20 mg/dL standard stock solution: Tabulate volume of solute and solvent and also dilution. ( 5 pts. each)

A. 2 mg/dL C. 8 mg/dL
B. 4 mg/dL D. 10 mg/dL
E. 15 mg/dL

ANSWER TO QUESTION NO. 2

What is the Molarity of a 0.5 N H2SO4 solution?

Since the Normality is known = 0.5 N

All you have to do is to make use of the shorter formula which is

M = Normality divided by the valence or N/v (factor)

M = 0.5 / 2

M = 0.25 M of H2SO4 (ANSWER)

ANSWER TO QUESTION NO. 3

What is the dilution if you add 0.75 ml of serum to 3.0 ml of diluent?

Consider the volume of serum as 1 part.

The dilution is the number of parts of the solute in proportion to the diluents plus 1.

So if 1 part = 0.75 mL

Then 3 ml is = 4 parts ( 3 ml divided by 0.75 ml)

So 1 part serum = 0.75 ml

+ 4 parts diluents = 3.0 ml

Will give you 5 parts ( 1 part serum + 4 parts diluent = 5 parts)

Dilution then is

1:5 (ANSWER)

QUESTION NO. 4

How would you prepare 1:20 serum dilution?

You can assume several volumes in this question. You can assume 1 ml as the volume of the solute as it is the easiest.

1. So 1 ml of serum + 19 ml of diluent = 1:20 serum dilution.

2. You can also assume a smaller volume for serum, especially if you were not able to extract more blood. Let’s say you have 0.5 ml of serum and you have to prepare 1:20 dilution. All you have to do is to assume that 0.5 ml is one part.

3. If 0.5 ml is one part, and you need 19 parts of the diluents to prepare a 1;20 dilution; all you have to do is to multiply 19 parts with 0.5 ml to get the volume of the 19 parts.

4. Your answer would then be 9.5 ml. ( 19 X 0.5 = 9.5)

5. So 0.5 ml of serum + 9.5 ml of diluent = 1:20 serum dilution (ANSWER)

QUESTION NO. 5

How would you prepare a 0.5 % NaCl solution?

The formula for % (W/V) is

% = W divided by the total volume multiplied by 100.

Therefore:

0.5% = W/V X 100

Assume that the total volume is 100 ml if no volume is given, so by transposing we come up with this formula:

W = 0.5% X 100 (total volume) / 100 %

W = 0.5 grams per 100 ml of diluent (ANSWER)

Note:

Convert volumes to ml and weights to grams to facilitate computation.

For QUESTION NUMBER 6, the ANSWERS WILL BE GIVEN NEXT POST.

Bilirubin Review Exam Questions

INDICATE WHETHER:

INCREASE B1, INCREASE TB = A

INCREASE B2, INCREASE TB = B

INCREASE B1 & B2, INCREASE TB = C

NONE OF THE ABOVE = D

1. 1. Obstruction of the biliary tract

2. Hepatic encephalopathy
3. EBF
4. Infectious Hepatitis
5. HDN
6. Liver carcinoma
7. Gilbert’s syndrome
8. Reye’s disease
9. Hepatitis B infection
10. Kernicterus

Where:

TB- TOTAL BILIRUBIN

B1 – Bilirubin 1

B2 – Bilirubin 2

ANSWERS FOUND IN THIS POST.

Answers to Normality, Molarity Problem Solving Review Questions in Clinical Chemistry

PROBLEM SOLVING :

1. WRITE DOWN FORMULA.
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 5 g NaCl solution and dissolve it in 1000 ml of diluent. ( 5 pts. each)

A. What is the Molarity?

M= Gram molecular weight per liter of solution, hence: M= 5 /58.5/1 liter

M = 0.085 M

B. What is the Normality?

Since Normality (N) = M x valence(factor)

M = 0.085 x 1 (valence)

M = 0.085 N

N.B. WHEN THE VALENCE OR FACTOR IS ONE THE NORMALITY IS EQUIVALENT TO THE MOLARITY.

C. What is the percent solution?

% = Weight/total volume x 100

% = (5/1,000 mL) X 100

% = 0.5%

ANSWERS TO THE REMAINING QUESTIONS WILL BE GIVEN NEXT POST.

2. What is the Molarity of a 0.5 N H2SO4 solution? (5 pts.)
3. What is the dilution if you add 0.75 ml of serum to 3.0 ml of diluent? ( 5 pts.)
4. How would you prepare 1:20 serum dilution? (5 pts.)
5. How would you prepare a 0.5 % NaCl solution? (5 pts.)
6. You are asked to prepare 200 ml of the following standard concentrations from a 20 mg/dL standard stock solution: Tabulate volume of solute and solvent and also dilution. ( 5 pts. each)

A. 2 mg/dL C. 8 mg/dL
B. 4 mg/dL D. 10 mg/dL
E. 15 mg/dL

Answers to Clinical Chemistry Review Questions - Blood Gas Analysis

Clinical Chemistry Review Questions vary in difficulty and complexity. Here are some basic questions which can help you review common principles in Clinical Chemistry.

Select the BEST ANSWER.

1. The most common sample specimen in clinical chemistry is:

a. Plasma c. whole blood

b. Serum d. buffy coat

2. In enzyme analysis, the following should be monitored closely, EXCEPT:

a. Temperature c. pH

b. Concentration of substrate d. non-competitive inhibitor

3. Electrolytes are called amphoteric substances because of this reason:

a. They can either be negatively or positively charged

b. They can be water or non-water soluble

c. They can transform from one energy form to another

d. They are directly transported in the blood stream.

4. The following statements are true of electrophoresis, EXCEPT:

a. It is a method to separate proteins from one another

b. The media can be paper, agar gel or cellulose

c. The principle depends upon their ability to fluoresce

d. Electrophoretic mobility is based on the charges of the ions.

5. In the maintenance of normal blood pH, these two organs are involved:

a. Lungs and heart c. lungs and kidneys

b. Kidneys and heart d. kidneys and liver

Choices for numbers 6 to 10

a. Uncompensated metabolic alkalosis

b. Uncompensated metabolic acidosis

c. Uncompensated respiratory alkalosis

d. Uncompensated respiratory acidosis

e. Partially compensated metabolic alkalosis

f. Partially compensated metabolic acidosis

g. Partially compensated respiratory alkalosis

h. Partially compensated respiratory acidosis

i. Fully compensated metabolic alkalosis

j. Fully compensated metabolic acidosis

k. Fully compensated respiratory alkalosis

l. Fully compensated respiratory acidosis

m. None of the above

6. Given pH – 7.49, pCO2= 40 mmHg, HCO3 = 32 mmol/L, What do these values indicate?- partially compensated metabolic alkalosis

7. Given pH – 6.8, dCO2= 10 mmol/L, TCO2= 22 mmol/L, What do these values indicate?- partially compensated respiratory acidosis

8. Given pH- 7.43, HCO3= 29 mmol/L, pCO2 – 50 mmHg. What do these values indicate?- Fully compensated metabolic alkalosis

-

9. Given pH- 7.15, HCO3= 25 mmol/L, pCO2 – 60 mmHg. What do these values indicate? - Uncompensated respiratory acidosis.

-

U10. Given pH- 7.50, HCO3= 40 mmol/L, pCO2 – 40 mmHg. What do these values indicate?

- Uncompensated metabolic alkalosis

Clinical Chemistry Review Questions

Clinical Chemistry Review Questions vary in difficulty and complexity. Here are some basic questions which can help you review common principles in Clinical Chemistry.

Select the BEST ANSWER.

1. The most common sample specimen in clinical chemistry is:

a. Plasma c. whole blood

b. Serum d. buffy coat

2. In enzyme analysis, the following should be monitored closely, EXCEPT:

a. Temperature c. pH

b. Concentration of substrate d. non-competitive inhibitor

3. Electrolytes are called amphoteric substances because of this reason:

a. They can either be negatively or positively charged

b. They can be water or non-water soluble

c. They can transform from one energy form to another

d. They are directly transported in the blood stream.

4. The following statements are true of electrophoresis, EXCEPT:

a. It is a method to separate proteins from one another

b. The media can be paper, agar gel or cellulose

c. The principle depends upon their ability to fluoresce

d. Electrophoretic mobility is based on the charges of the ions.

5. In the maintenance of normal blood pH, these two organs are involved:

a. Lungs and heart c. lungs and kidneys

b. Kidneys and heart d. kidneys and liver

Choices for numbers 6 to 10

a. Uncompensated metabolic alkalosis

b. Uncompensated metabolic acidosis

c. Uncompensated respiratory alkalosis

d. Uncompensated respiratory acidosis

e. Partially compensated metabolic alkalosis

f. Partially compensated metabolic acidosis

g. Partially compensated respiratory alkalosis

h. Partially compensated respiratory acidosis

i. Fully compensated metabolic alkalosis

j. Fully compensated metabolic acidosis

k. Fully compensated respiratory alkalosis

l. Fully compensated respiratory acidosis

m. None of the above

6. Given pH – 7.49, pCO2= 40 mmHg, HCO3 = 32 mmol/L, What do these values indicate?

7. Given pH – 6.8, dCO2= 10 mmol/L, TCO2= 22 mmol/L, What do these values indicate?

8. Given pH- 7.43, HCO3= 29 mmol/L, pCO2 – 50 mmHg. What do these values indicate?

9. Given pH- 7.15, HCO3= 25 mmol/L, pCO2 – 60 mmHg. What do these values indicate?

10. Given pH- 7.50, HCO3= 40 mmol/L, pCO2 – 40 mmHg. What do these values indicate?