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Laboratory Operations: Lab math – page 3

48 Laboratory Operations MCQs on Lab math with answers and explanations.

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Q41MediumLab math

How many grams of sodium chloride (MW 58.5 g/mol) are needed to prepare 500 mL of a 0.1 mol/L solution?

Answer: A. 2.925 g

Moles needed = 0.1 × 0.5 = 0.05 mol, and 0.05 × 58.5 = 2.925 g.

ID MG-ELOP-0032 · Found a mistake? Report it
Q42HardLab math

In a kinetic assay, NADH (molar absorptivity 6220 L·mol⁻¹·cm⁻¹) gives an absorbance of 0.311 in a 1 cm cuvette. The NADH concentration is:

Answer: C. 50 µmol/L

c = A/(εb) = 0.311/(6220 × 1) = 5.0 × 10⁻⁵ mol/L = 50 µmol/L. Check the power of ten carefully when converting to µmol/L.

ID MG-LOP-0036 · Found a mistake? Report it
Q43HardLab math

Concentrated hydrochloric acid is 37% (w/w) HCl with a specific gravity of 1.19. Given MW 36.5 g/mol, its approximate molarity is:

Answer: C. 12.1 mol/L

Molarity = (SG × 1000 × % purity)/MW = (1.19 × 1000 × 0.37)/36.5 ≈ 12.1 mol/L. Forgetting the specific gravity gives 10.1 mol/L.

ID MG-LOP-0041 · Found a mistake? Report it
Q44HardLab math

Arterial blood has bicarbonate 24 mmol/L and pCO2 40 mm Hg. Using pH = 6.1 + log [HCO3⁻/(0.03 × pCO2)], the pH is:

Answer: C. 7.40

0.03 × 40 = 1.2 mmol/L; 24/1.2 = 20; log 20 = 1.30; 6.1 + 1.30 = 7.40.

ID MG-LOP-0044 · Found a mistake? Report it
Q45HardLab math

How many grams of calcium chloride (CaCl2, MW 111 g/mol) are needed to prepare 1 L of a solution containing 100 mEq/L?

Answer: B. 5.55 g

Ca²⁺ has valence 2, so 100 mEq = 50 mmol. 50 mmol × 111 mg/mmol = 5550 mg = 5.55 g. 11.1 g treats mEq as mmol.

ID MG-LOP-0046 · Found a mistake? Report it
Q46HardLab math

A WBC count uses a 1:20 dilution. A total of 100 cells are counted in the four large corner squares (each 1 mm², depth 0.1 mm). The WBC count is:

Answer: B. 5.0 × 10⁹/L

Volume counted = 4 × 1 × 0.1 = 0.4 µL. Cells/µL = (100 × 20)/0.4 = 5000/µL = 5.0 × 10⁹/L.

ID MG-LOP-0047 · Found a mistake? Report it
Q47HardLab math

A procedure requires 1000 × g. The centrifuge rotor radius is 15 cm. Using RCF = 1.118 × 10⁻⁵ × r × rpm², the required speed is about:

Answer: B. 2440 rpm

rpm = √[1000 ÷ (1.118 × 10⁻⁵ × 15)] = √(1000 ÷ 0.0001677) = √5,963,000 ≈ 2440 rpm. The value 5960 comes from not taking the square root correctly.

ID MG-LOP-0127 · Found a mistake? Report it
Q48HardLab math

A creatinine clearance is 100 mL/min in a patient with a body surface area of 1.20 m². Corrected to 1.73 m², the clearance is about:

Answer: C. 144 mL/min/1.73 m²

Corrected clearance = 100 × 1.73/1.20 ≈ 144 mL/min/1.73 m². Multiplying by 1.20/1.73 instead gives 69.

ID MG-LOP-0130 · Found a mistake? Report it
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