MLT GlobeLaboratory Operations: Lab math
48 Laboratory Operations MCQs on Lab math with answers and explanations.
How much sodium hydroxide (MW 40) is needed to prepare 1 litre of N/5 (0.2 N) NaOH?
For NaOH the equivalent weight equals 40 g. 0.2 N × 40 g/L = 8.0 g per litre. 4 g would give 0.1 N.
A serum is diluted 1:10, and that dilution is then diluted 1:5. The final dilution is:
Serial dilutions multiply: 1/10 × 1/5 = 1/50. Multiply the measured result by 50 to get the original concentration.
How many grams of sodium chloride are needed to prepare 500 mL of a 0.9% (w/v) saline solution?
0.9% w/v means 0.9 g per 100 mL. For 500 mL: 0.9 × 5 = 4.5 g. 9.0 g would be needed for 1 L, not 500 mL.
What is the normality of a 0.5 mol/L sulfuric acid (H2SO4) solution?
H2SO4 gives 2 replaceable H+ per molecule, so N = M × 2 = 0.5 × 2 = 1.0 N. Using 0.5 N ignores the valence of the acid.
How many millilitres of a 10% stock solution are needed to prepare 200 mL of a 2% solution?
Using C1V1 = C2V2: 10% × V1 = 2% × 200 mL, so V1 = 40 mL, made up to 200 mL with diluent.
A 100 mg/dL standard gives an absorbance of 0.25. A patient sample treated the same way gives 0.40. Assuming Beer's law is obeyed, the patient concentration is:
Cu = (Au/As) × Cs = (0.40/0.25) × 100 = 160 mg/dL. 62.5 mg/dL results from inverting the ratio.
A fasting glucose result is 90 mg/dL. Given a molecular weight of 180 g/mol, what is the result in SI units?
90 mg/dL = 900 mg/L; 900 ÷ 180 = 5.0 mmol/L (conversion factor 0.0555). 0.5 mmol/L forgets to convert dL to L.
A freezer used for storing sera must be kept at −20 °C. What is this temperature in degrees Fahrenheit?
°F = (°C × 1.8) + 32 = (−20 × 1.8) + 32 = −36 + 32 = −4 °F. −36 °F forgets to add 32.
A stock standard contains 1000 mg/dL glucose. What volume of stock is needed to prepare 10 mL of a 50 mg/dL working standard?
V1 = (C2 × V2)/C1 = (50 × 10)/1000 = 0.5 mL, then dilute to 10 mL. 5.0 mL would give 500 mg/dL.
A total cholesterol is 200 mg/dL. Cholesterol has a molecular weight of 386.7 g/mol. The result in SI units is about:
200 mg/dL = 2000 mg/L; 2000 ÷ 386.7 = 5.17 mmol/L. Forgetting to convert dL to L gives 0.52 mmol/L.
Serum results are sodium 138 mmol/L, chloride 102 mmol/L and bicarbonate 24 mmol/L. Using Na⁺ − (Cl⁻ + HCO₃⁻), the anion gap is:
Anion gap = 138 − (102 + 24) = 12 mmol/L. Adding potassium would give about 16 mmol/L, but this formula does not include potassium.
A blood sample has a haemoglobin of 15 g/dL and a haematocrit of 45%. The MCHC is:
MCHC = Hb ÷ Hct × 100 = 15/45 × 100 = 33.3 g/dL, within the normal range. Dividing Hct by Hb (3.0) inverts the formula.
To make a 1:10 dilution using 0.2 mL of serum, what should the total final volume be?
In a 1:10 dilution the sample is one tenth of the total volume, so 0.2 mL × 10 = 2.0 mL total (0.2 mL serum + 1.8 mL diluent).
How many microlitres are there in 5 mL?
1 mL = 1000 µL, so 5 mL = 5000 µL.
5 g of a solute is dissolved in water to a final volume of 250 mL. What is the percent (w/v) concentration?
Percent w/v is grams per 100 mL: 5 g ÷ 250 mL × 100 = 2%.
A haemoglobin result of 14 g/dL expressed in g/L is:
1 L = 10 dL, so multiply by 10: 14 g/dL = 140 g/L.
A technologist dissolves 5.85 g of NaCl (molecular weight 58.5 g/mol) in water to a final volume of 250 mL. What is the molarity?
Moles = 5.85/58.5 = 0.1 mol. Molarity = 0.1 mol ÷ 0.25 L = 0.4 mol/L. 0.1 mol/L forgets to divide by the volume in litres.
A doubling serial dilution is set up; tube 1 contains a 1:2 dilution of patient serum. The last tube showing agglutination is tube 6. What is the titer?
In a twofold series starting at 1:2, tube n = 1:2^n. Tube 6 = 1:64. 1:32 is tube 5.
A solution reads 25% transmittance on a spectrophotometer. What is its absorbance?
A = 2 − log %T = 2 − log 25 = 2 − 1.398 = 0.602. Absorbance is not simply 1 − T (0.75); the relationship is logarithmic.
A centrifuge has a rotor radius of 10 cm and runs at 3000 rpm. Using RCF = 1.118 × 10⁻⁵ × r × rpm², the relative centrifugal force is about:
RCF = 1.118 × 10⁻⁵ × 10 × (3000)² = 1.118 × 10⁻⁵ × 9 × 10⁷ ≈ 1006 × g. rpm alone (3000) is not the same as g-force.