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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 example

How 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.

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Milliequivalents
How many milliequivalents are in 2 g of potassium chloride (KCl)? (MW = 74.5, valence = 1) Round your answer to one decimal place.
mEq

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How to set up mmol, mEq and mOsm problems

These three units count different things about the same salt. Millimoles count molecules. Milliequivalents count charges, which is what matters for electrolyte replacement, because a potassium ion does the same job whether it came from KCl or potassium acetate. Milliosmoles count particles in solution, which is what matters for tonicity, because osmotic pressure depends on how many things are dissolved, not what they are. All three start from the same first step: grams to millimoles using the molecular weight.

mmol = mg ÷ MW   •   mEq = mmol × valence   •   mOsm = mmol × number of particles the salt dissociates into

The molecular weight is in mg per mmol (which is the same number as g per mol), so the mass has to be in mg before dividing. The exam gives you the molecular weight and, usually, the valence. It does not always tell you how many particles a salt dissociates into; that you are expected to see from the formula.

Worked example 1: millimoles

How many millimoles are in 3 g of sodium chloride (MW 58.5)?

mg = 3 × 1,000 = 3,000 mg

mmol = 3,000 ÷ 58.5 = 51.28 → 51.3 mmol

Worked example 2: milliequivalents of a divalent salt

How many milliequivalents are in 2 g of calcium chloride dihydrate (MW 147, valence 2)?

mmol = 2,000 ÷ 147 = 13.605 mmol

mEq = 13.605 × 2 = 27.21 → 27.2 mEq

The valence is the charge on the ion you care about. Calcium is Ca²⁺, so each millimole carries two milliequivalents. For sodium, potassium and chloride the valence is 1 and mEq equals mmol, which is why people forget the step exists until a calcium or magnesium question shows up. Use the molecular weight of the salt as supplied, including any water of hydration: the dihydrate weighs 147, not 111, and the exam will give you the number for the form in the question.

Worked example 3: osmolarity

What is the osmolarity of 0.9% sodium chloride in mOsm/L? (MW 58.5; NaCl dissociates into 2 particles.)

grams per litre = 0.9% × 1,000 = 9 g

mOsm/L = 9 ÷ 58.5 × 2 × 1,000 = 307.69 → 308 mOsm/L

Work in grams per litre, divide by the molecular weight to get moles per litre, multiply by the particles, and multiply by 1,000 to turn osmoles into milliosmoles. The 1,000 at the end is the one that gets dropped. 308 mOsm/L is the figure printed on a bag of normal saline, and knowing it lets you check your own arithmetic.

The same setup for 5% dextrose uses dextrose monohydrate, MW 198, and one particle because sugar doesn't dissociate:

mOsm/L = 50 ÷ 198 × 1 × 1,000 = 252.5 mOsm/L

The bag says 252. If you use the anhydrous molecular weight of 180 you get 278, which matches no bag on the shelf. The molecular weight in the question is the one to use.

Counting particles

Valence and particle count are different numbers that happen to coincide for the monovalent salts. CaCl₂ has valence 2 for the calcium but dissociates into 3 particles. MgSO₄ has valence 2 and 2 particles. Keep the two questions separate: "how many charges" is mEq, "how many pieces" is mOsm.

Where students lose the points

10 Milliequivalent (mEq), Millimole & Milliosmole practice problems with worked solutions

Every problem below is worked start to finish. Try it yourself first, then check the steps. For a fresh problem every time, use the practice box above. It generates unlimited variations of these same question types.

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
  1. Step 1: grams in 1,000 mL: 5% = 5 g/100 mL, so in 1,000 mL = 50 g.
  2. Step 2: apply mOsm/L = (mass in g ÷ MW) × number of particles × 1,000.
  3. = (50 g/L ÷ 198 g/mol) × 1 particles × 1,000 mOsm/osmol ≈ 252.53 mOsm/L.
  4. 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
  1. Convert to mg: 14.5 g = 14500 mg.
  2. mmol = mg ÷ MW = 14500 mg ÷ 147 mg/mmol ≈ 98.64 mmol.
  3. 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
  1. mg: 4.5 g = 4500 mg.
  2. mmol = 4500 mg ÷ 84 mg/mmol ≈ 53.571 mmol.
  3. mEq = mmol × valence = 53.571 mmol × 1 mEq/mmol ≈ 53.57 mEq.
  4. 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
  1. Step 1: grams in 1,000 mL: 0.3% = 0.3 g/100 mL, so in 1,000 mL = 3 g.
  2. Step 2: apply mOsm/L = (mass in g ÷ MW) × number of particles × 1,000.
  3. = (3 g/L ÷ 74.5 g/mol) × 2 particles × 1,000 mOsm/osmol ≈ 80.54 mOsm/L.
  4. 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
  1. Convert to mg: 17.75 g = 17750 mg.
  2. mmol = mg ÷ MW = 17750 mg ÷ 147 mg/mmol ≈ 120.75 mmol.
  3. 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
  1. mg: 4.5 g = 4500 mg.
  2. mmol = 4500 mg ÷ 58.5 mg/mmol ≈ 76.923 mmol.
  3. mEq = mmol × valence = 76.923 mmol × 1 mEq/mmol ≈ 76.92 mEq.
  4. 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
  1. Step 1: grams in 1,000 mL: 4.2% = 4.2 g/100 mL, so in 1,000 mL = 42 g.
  2. Step 2: apply mOsm/L = (mass in g ÷ MW) × number of particles × 1,000.
  3. = (42 g/L ÷ 84 g/mol) × 2 particles × 1,000 mOsm/osmol = 1000 mOsm/L.
  4. 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
  1. Convert to mg: 14 g = 14000 mg.
  2. mmol = mg ÷ MW = 14000 mg ÷ 246.5 mg/mmol ≈ 56.8 mmol.
  3. 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
  1. mg: 0.5 g = 500 mg.
  2. mmol = 500 mg ÷ 74.5 mg/mmol ≈ 6.711 mmol.
  3. mEq = mmol × valence = 6.711 mmol × 1 mEq/mmol ≈ 6.71 mEq.
  4. 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
  1. Step 1: grams in 1,000 mL: 0.9% = 0.9 g/100 mL, so in 1,000 mL = 9 g.
  2. Step 2: apply mOsm/L = (mass in g ÷ MW) × number of particles × 1,000.
  3. = (9 g/L ÷ 58.5 g/mol) × 2 particles × 1,000 mOsm/osmol ≈ 307.69 mOsm/L.
  4. 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).

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