Dilutions, concentrations and solution preparation

13 DOH Abu Dhabi (HAAD) lab licensing practice questions on Dilutions, concentrations and solution preparation, part of General Laboratory, Quality and Safety. The first 13 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Hard

   To confirm a borderline meropenem MIC on a Klebsiella pneumoniae isolate, a reference laboratory prepares its own broth microdilution panel by the CLSI method. It needs 20 mL of a meropenem stock solution at 1,280 µg/mL, which will later be diluted 1 in 10 to give a 128 µg/mL working solution. The certificate for the reference powder lot gives an assay potency of 880 µg of active drug per mg of powder. The mass of powder to weigh for the stock solution is:

   1. A25.6 mg
   2. B2.9 mg
   3. C29.1 mg
   4. D22.5 mg

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   Answer: C · 29.1 mg

   CLSI gives weight (mg) = volume (mL) × concentration (µg/mL) / potency (µg/mg). Here 20 × 1,280 = 25,600 µg of active drug is needed, and each mg of powder supplies only 880 µg, so 25,600 / 880 = 29.1 mg of powder is weighed. Ignoring potency gives 25.6 mg, a stock about 12% weak, which would make the MICs read falsely high. The 1 in 10 working dilution is made from the stock afterwards and does not change the weighing.

   • A: This ignores potency: each mg of powder holds only 880 µg of drug.
   • B: The 1 in 10 working dilution is made from the stock later; it does not change the stock weighing.
   • D: This multiplies by the potency fraction instead of dividing by it, making the stock even weaker.

2. Question 2 · Hard

   A 24-year-old woman, 11 weeks after her last period, has heavy vaginal bleeding, and ultrasound suggests a molar pregnancy. Her serum hCG exceeds the analyzer's measuring range of up to 10,000 IU/L even with the automatic 1:10 on-board dilution, which the analyzer corrects for itself. The technologist makes a manual dilution of 50 µL of serum in 950 µL of the manufacturer's diluent and loads it, again with the 1:10 on-board dilution selected. The analyzer reports 8,450 IU/L for this manual dilution. The hCG result to report is:

   1. A169,000 IU/L
   2. B8,450 IU/L
   3. C160,550 IU/L
   4. D1,690,000 IU/L

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   Answer: A · 169,000 IU/L

   Mixing 50 µL of serum with 950 µL of diluent gives 50 µL in 1,000 µL, a 1 in 20 dilution, so the factor is 20, not 19. The analyzer has already multiplied its reading by 10 for the on-board dilution, so only the manual factor remains: 8,450 × 20 = 169,000 IU/L. Multiplying by 10 again counts the on-board dilution twice, and reporting 8,450 IU/L ignores the manual dilution. Manual dilutions use the manufacturer's diluent so the matrix matches, and an hCG this high fits a molar pregnancy.

   • B: This is the result of the diluted sample, not of the patient's serum.
   • C: 50 µL in 950 µL is a 1 in 20 dilution (50 / 1,000), not 1 in 19.
   • D: This applies the on-board factor a second time; the analyzer has already corrected for it.

3. Question 3 · Hard

   A glucose stock standard contains 10.0 g/L. How much stock is needed to prepare 100 mL of a 5.0 mmol/L working standard (glucose molar mass 180 g/mol)?

   1. A50 mL
   2. B0.28 mL
   3. C0.90 mL
   4. D9.0 mL

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   Answer: D · 9.0 mL

   First put both concentrations in the same units: 5.0 mmol/L x 180 mg/mmol = 900 mg/L, or 0.90 g/L. Then C1 x V1 = C2 x V2: 10.0 g/L x V1 = 0.90 g/L x 100 mL, so V1 = 9.0 mL, made up to 100 mL. Treating 5.0 mmol/L as if it were 5.0 g/L gives 50 mL, a common error. Units must match on both sides before C1V1 is used.

   • A: 50 mL treats 5.0 mmol/L as 5.0 g/L, without converting units.
   • B: 0.28 mL divides by the molar mass instead of multiplying by it.
   • C: 0.90 mL comes from a tenfold slip when converting between mg and g.

4. Question 4 · Hard

   A 2.0 mmol/L stock is diluted 1 in 20, and this mixture is then diluted 1 in 25 to make a low calibrator. What is the calibrator's concentration?

   1. A0.40 µmol/L
   2. B80 µmol/L
   3. C44 µmol/L
   4. D4.0 µmol/L

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   Answer: D · 4.0 µmol/L

   When dilutions are made in series, the overall dilution is the product of the steps: 20 x 25 = 500. The final concentration is 2.0 mmol/L / 500 = 0.0040 mmol/L, which is 4.0 µmol/L. Adding the dilution factors (20 + 25 = 45) instead gives about 44 µmol/L, a common error. Two-step dilutions are used when a single step would need a volume too small to pipette accurately.

   • A: 0.40 µmol/L is a tenfold slip in converting mmol/L to µmol/L.
   • B: 80 µmol/L applies only the 1 in 25 step.
   • C: 44 µmol/L divides by the sum of the dilution factors (45) instead of their product.

5. Question 5 · Hard

   A bacterial suspension is diluted in tenfold steps from 10^-1 to 10^-6. Plating 0.1 mL of the 10^-5 dilution gives 48 colonies. What is the count in the original suspension?

   1. A4.8 x 10^6 CFU/mL
   2. B4.8 x 10^5 CFU/mL
   3. C4.8 x 10^7 CFU/mL
   4. D4.8 x 10^8 CFU/mL

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   Answer: C · 4.8 x 10^7 CFU/mL

   The plate count is first converted to colonies per mL of the dilution plated: 48 / 0.1 mL = 480 CFU/mL. This is then multiplied by the dilution factor, 10^5, giving 4.8 x 10^7 CFU/mL. Forgetting the 0.1 mL plating volume gives a tenfold underestimate. Counts are most reliable from plates with about 30-300 colonies.

   • A: 4.8 x 10^6 forgets to correct for plating only 0.1 mL.
   • B: 4.8 x 10^5 multiplies by the 0.1 mL plated instead of dividing by it.
   • D: 4.8 x 10^8 uses the factor of the 10^-6 tube instead of the 10^-5 tube.

6. Question 6 · Medium

   A uric acid result from another hospital is 7.0 mg/dL. Your laboratory reports uric acid in µmol/L (molar mass 168 g/mol). The equivalent result is about:

   1. A41.7 µmol/L
   2. B417 µmol/L
   3. C24 µmol/L
   4. D4,170 µmol/L

   Answer and explanation in LearnFree account

7. Question 7 · Hard

   A method needs 250 mL of 0.20 N sulfuric acid (H2SO4, molar mass 98 g/mol). How many grams of pure H2SO4 does this volume contain?

   1. A4.90 g
   2. B2.45 g
   3. C1.23 g
   4. D9.80 g

   Answer and explanation in LearnFree account

8. Question 8 · Medium

   How many grams of sodium hydroxide (NaOH, molar mass 40.0 g/mol) are needed for 250 mL of a 0.50 mol/L solution?

   1. A5.0 g
   2. B10 g
   3. C0.125 g
   4. D20 g

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9. Question 9 · Hard

   A free T4 of 1.2 ng/dL on a referral report must be converted to pmol/L (thyroxine molar mass 777 g/mol). The equivalent result is about:

   1. A0.0154 pmol/L
   2. B15.4 pmol/L
   3. C154 pmol/L
   4. D1.54 pmol/L

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10. Question 10 · Easy

    A reagent labeled 5% (w/v) contains:

    1. A5 g of solute per 100 g of final solution
    2. B5 mL of solute per 100 mL of final solution
    3. C5 g of solute added to 100 mL of water
    4. D5 g of solute per 100 mL of final solution

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11. Question 11 · Hard

    A patient transferred to an Abu Dhabi hospital brings a report showing a blood urea nitrogen (BUN) of 28 mg/dL. The laboratory reports urea in mmol/L. What is the equivalent urea result?

    1. A21.4 mmol/L
    2. B4.7 mmol/L
    3. C10.0 mmol/L
    4. D1.0 mmol/L

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12. Question 12 · Medium

    Household bleach containing 5% sodium hypochlorite is used to make 1 L of 0.5% hypochlorite for cleaning blood spills. The correct mixture is:

    1. A100 mL of bleach plus 1,000 mL of water
    2. B10 mL of bleach plus 990 mL of water
    3. C100 mL of bleach plus 900 mL of water
    4. D50 mL of bleach plus 950 mL of water

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13. Question 13 · Medium

    A glucose result is 180 mg/dL. What is it in SI units (glucose molar mass about 180 g/mol)?

    1. A3.24 mmol/L
    2. B10.0 mmol/L
    3. C1.0 mmol/L
    4. D18.0 mmol/L

    Answer and explanation in LearnFree account