Laboratory Mathematics and Calculations Flashcards

A specimen undergoes serial dilutions of 1:2, then 1:5, then 1:10. What is the final dilution?

Answer: 1:100

Serial dilutions multiply: 2 × 5 × 10 = 100, giving a 1:100 final dilution.

How many grams of NaCl (MW 58.5 g/mol) are needed to prepare 500 mL of a 0.2 M solution?

Answer: 5.85 g

0.2 mol/L × 0.5 L = 0.1 mol, and 0.1 mol × 58.5 g/mol = 5.85 g.

What is the normality of a 1 M sulfuric acid (H2SO4) solution?

Answer: 2 N

H2SO4 donates two H+ ions per molecule, so normality equals 2 × molarity.

How much 2 M stock solution is needed to prepare 100 mL of a 0.5 M working solution?

Answer: 25 mL

Using C1V1 = C2V2: V1 = (0.5 × 100) / 2 = 25 mL.

10 g of solute is dissolved in water to a final volume of 250 mL. What is the % w/v concentration?

Answer: 4%

% w/v is grams per 100 mL: 10 g / 250 mL × 100 = 4%.

A glucose result is 180 mg/dL (MW 180 g/mol). What is the equivalent in mmol/L?

Answer: 10.0 mmol/L

180 mg/dL = 1800 mg/L, and 1800 mg/L ÷ 180 mg/mmol = 10 mmol/L.

A specimen diluted 1:4 gives a reading of 75 mg/dL. What result should be reported?

Answer: 300 mg/dL

Multiply the diluted reading by the dilution factor: 75 × 4 = 300 mg/dL.