Milliequivalent (mEq), Millimole & Milliosmole Practice
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These three units describe amounts of electrolytes. Millimoles (mmol) measure the number of molecules, milliequivalents (mEq) account for ionic charge (valence), and milliosmoles (mOsm) count the total dissociated particles that determine osmotic pressure.
Formulasmmol = mg ÷ molecular weight • mEq = mmol × valencemOsm/L = (grams per liter ÷ MW) × number of particles × 1,000
Worked exampleHow many mEq are in 1 g of KCl (MW 74.5, valence 1)? mmol = 1,000 ÷ 74.5 = 13.4; mEq = 13.4 × 1 = 13.4 mEq. Osmolarity of 0.9% NaCl = 9 g/L ÷ 58.5 × 2 × 1,000 = 308 mOsm/L.
Valence trap: monovalent ions (Na⁺, K⁺) → mEq = mmol; divalent ions (Ca²⁺, Mg²⁺) → 2 mEq per mmol.
10 Milliequivalent (mEq), Millimole & Milliosmole practice problems with worked solutions
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Problem 1: Milliosmoles
Calculate the osmolarity of a 5% dextrose monohydrate solution in mOsm/L. (MW = 198; does not dissociate (1 particle per molecule).) Round your answer to the nearest whole number.
Answer 253 mOsm/L
Solution- Step 1: grams in 1,000 mL: 5% = 5 g/100 mL, so in 1,000 mL = 50 g.
- Step 2: apply mOsm/L = (mass in g ÷ MW) × number of particles × 1,000.
- = (50 g/L ÷ 198 g/mol) × 1 particles × 1,000 mOsm/osmol ≈ 252.53 mOsm/L.
- Rounded as the question asks: 253 mOsm/L.
Landmarks: NS ≈ 308 mOsm/L, D5W ≈ 252 mOsm/L (dextrose monohydrate, MW 198). Both close enough to plasma (about 285 to 295) to run peripherally. Particles: NaCl = 2, dextrose = 1 (nonelectrolyte).
Problem 2: Millimoles
How many millimoles are in 14.5 g of calcium chloride dihydrate (CaCl₂·2H₂O)? (MW = 147) Round your answer to one decimal place.
Answer 98.6 mmol
Solution- Convert to mg: 14.5 g = 14500 mg.
- mmol = mg ÷ MW = 14500 mg ÷ 147 mg/mmol ≈ 98.64 mmol.
- Rounded as the question asks: 98.6 mmol.
Problem 3: Milliequivalents
How many milliequivalents are in 4.5 g of sodium bicarbonate (NaHCO₃)? (MW = 84, valence = 1) Round your answer to one decimal place.
Answer 53.6 mEq
Solution- mg: 4.5 g = 4500 mg.
- mmol = 4500 mg ÷ 84 mg/mmol ≈ 53.571 mmol.
- mEq = mmol × valence = 53.571 mmol × 1 mEq/mmol ≈ 53.57 mEq.
- Rounded as the question asks: 53.6 mEq.
Monovalent ions: mEq = mmol. Divalent (Ca²⁺, Mg²⁺): 2 mEq per mmol. Forgetting valence is the classic error.
Problem 4: Milliosmoles
Calculate the osmolarity of a 0.3% potassium chloride (KCl) solution in mOsm/L. (MW = 74.5; dissociates into 2 particles per molecule.) Round your answer to the nearest whole number.
Answer 81 mOsm/L
Solution- Step 1: grams in 1,000 mL: 0.3% = 0.3 g/100 mL, so in 1,000 mL = 3 g.
- Step 2: apply mOsm/L = (mass in g ÷ MW) × number of particles × 1,000.
- = (3 g/L ÷ 74.5 g/mol) × 2 particles × 1,000 mOsm/osmol ≈ 80.54 mOsm/L.
- Rounded as the question asks: 81 mOsm/L.
Landmarks: NS ≈ 308 mOsm/L, D5W ≈ 252 mOsm/L (dextrose monohydrate, MW 198). Both close enough to plasma (about 285 to 295) to run peripherally. Particles: NaCl = 2, dextrose = 1 (nonelectrolyte).
Problem 5: Millimoles
How many millimoles are in 17.75 g of calcium chloride dihydrate (CaCl₂·2H₂O)? (MW = 147) Round your answer to one decimal place.
Answer 120.7 mmol
Solution- Convert to mg: 17.75 g = 17750 mg.
- mmol = mg ÷ MW = 17750 mg ÷ 147 mg/mmol ≈ 120.75 mmol.
- Rounded as the question asks: 120.7 mmol.
Problem 6: Milliequivalents
How many milliequivalents are in 4.5 g of sodium chloride (NaCl)? (MW = 58.5, valence = 1) Round your answer to one decimal place.
Answer 76.9 mEq
Solution- mg: 4.5 g = 4500 mg.
- mmol = 4500 mg ÷ 58.5 mg/mmol ≈ 76.923 mmol.
- mEq = mmol × valence = 76.923 mmol × 1 mEq/mmol ≈ 76.92 mEq.
- Rounded as the question asks: 76.9 mEq.
Monovalent ions: mEq = mmol. Divalent (Ca²⁺, Mg²⁺): 2 mEq per mmol. Forgetting valence is the classic error.
Problem 7: Milliosmoles
Calculate the osmolarity of a 4.2% sodium bicarbonate (NaHCO₃) solution in mOsm/L. (MW = 84; dissociates into 2 particles per molecule.) Round your answer to the nearest whole number.
Answer 1,000 mOsm/L
Solution- Step 1: grams in 1,000 mL: 4.2% = 4.2 g/100 mL, so in 1,000 mL = 42 g.
- Step 2: apply mOsm/L = (mass in g ÷ MW) × number of particles × 1,000.
- = (42 g/L ÷ 84 g/mol) × 2 particles × 1,000 mOsm/osmol = 1000 mOsm/L.
- Rounded as the question asks: 1,000 mOsm/L.
Landmarks: NS ≈ 308 mOsm/L, D5W ≈ 252 mOsm/L (dextrose monohydrate, MW 198). Both close enough to plasma (about 285 to 295) to run peripherally. Particles: NaCl = 2, dextrose = 1 (nonelectrolyte).
Problem 8: Millimoles
How many millimoles are in 14 g of magnesium sulfate heptahydrate (MgSO₄·7H₂O)? (MW = 246.5) Round your answer to one decimal place.
Answer 56.8 mmol
Solution- Convert to mg: 14 g = 14000 mg.
- mmol = mg ÷ MW = 14000 mg ÷ 246.5 mg/mmol ≈ 56.8 mmol.
- Rounded as the question asks: 56.8 mmol.
Problem 9: Milliequivalents
How many milliequivalents are in 0.5 g of potassium chloride (KCl)? (MW = 74.5, valence = 1) Round your answer to one decimal place.
Answer 6.7 mEq
Solution- mg: 0.5 g = 500 mg.
- mmol = 500 mg ÷ 74.5 mg/mmol ≈ 6.711 mmol.
- mEq = mmol × valence = 6.711 mmol × 1 mEq/mmol ≈ 6.71 mEq.
- Rounded as the question asks: 6.7 mEq.
Monovalent ions: mEq = mmol. Divalent (Ca²⁺, Mg²⁺): 2 mEq per mmol. Forgetting valence is the classic error.
Problem 10: Milliosmoles
Calculate the osmolarity of a 0.9% sodium chloride (NaCl) solution in mOsm/L. (MW = 58.5; dissociates into 2 particles per molecule.) Round your answer to the nearest whole number.
Answer 308 mOsm/L
Solution- Step 1: grams in 1,000 mL: 0.9% = 0.9 g/100 mL, so in 1,000 mL = 9 g.
- Step 2: apply mOsm/L = (mass in g ÷ MW) × number of particles × 1,000.
- = (9 g/L ÷ 58.5 g/mol) × 2 particles × 1,000 mOsm/osmol ≈ 307.69 mOsm/L.
- Rounded as the question asks: 308 mOsm/L.
Landmarks: NS ≈ 308 mOsm/L, D5W ≈ 252 mOsm/L (dextrose monohydrate, MW 198). Both close enough to plasma (about 285 to 295) to run peripherally. Particles: NaCl = 2, dextrose = 1 (nonelectrolyte).