Osmolar

Osmolarity practice problems

Thirty worked problems with full answers: molarity × i, serum osmolality, gaps, IV mixing, PN and ORS.

Thirty problems with fully worked answers, grouped by topic. Try each one before opening the answer.

Conventions used throughout (the same as Osmolar’s calculators):

  • glucose mg/dL ÷ 18 = mmol/L
  • BUN mg/dL ÷ 2.8 = urea mmol/L; urea mg/dL ÷ 6.0 = mmol/L
  • ethanol mg/dL ÷ 3.7 (default; ÷ 4.6 shown as the alternative)
  • Intermediates stay unrounded. Only the final displayed result is rounded, half away from zero.

Every answer is arithmetic for learning, not a clinical decision or a treatment instruction.

A. Molarity × i (solution osmolarity)

Ideal osmolarity (mOsm/L) = mmol/L × number of particles (i). Background: IV fluid osmolarity.

A1. What is the calculated osmolarity of 0.9% NaCl (9 g/L; NaCl 58.44 g/mol)?

Answer

9 ÷ 58.44 = 0.15400 mol/L = 154.0 mmol/L. NaCl gives 2 particles: 154.0 × 2 = 308 mOsm/L (calculated). This matches the label’s “308 mOsmol/L (calc)” [1].

A2. Calculate the osmolarity of 3% NaCl (30 g/L).

Answer

30 ÷ 58.44 = 0.51335 mol/L, and × 2 = 1,026.7 → 1,027 mOsm/L. The B. Braun label prints 1,030. That is a manufacturer rounding difference, not a different solution [2].

A3. D5W contains 50 g/L dextrose. What is its osmolarity using hydrous dextrose (198.17 g/mol) and using anhydrous dextrose (180.16 g/mol)? Which matches US labels?

Answer
  • Hydrous: 50 ÷ 198.17 × 1,000 = 252.3 → 252 mOsm/L
  • Anhydrous: 50 ÷ 180.16 × 1,000 = 277.5 → 278 mOsm/L

US IV labels use hydrous dextrose. Baxter prints 252 [3]. B. Braun prints 250 for the same 50 g/L, a manufacturer convention [4]. State which molar mass you used.

A4. Lactated Ringer’s (Baxter) lists Na⁺ 130, K⁺ 4, Ca²⁺ 2.7 mEq/L, Cl⁻ 109 and lactate 28 per litre. Calculate the osmolarity. What happens if calcium is entered as 2.7 mmol/L?

Answer

Ca²⁺ is divalent, so 2.7 mEq ÷ 2 = 1.35 mmol. 130 + 4 + 1.35 + 109 + 28 = 272.35 → 272 mOsm/L. The label prints 273, a rounding difference from the stated ions [5]. Entering calcium as 2.7 mmol gives 273.7 → 274, which is the mEq/mmol trap for divalent ions.

B. Osmolality vs osmolarity

Background: osmolality vs osmolarity.

B1. 0.9% NaCl is labelled 308 mOsm/L (calculated), but measures about 286 mOsm/kg [6]. What osmotic coefficient does that imply?

Answer

286 ÷ 308 = 0.93. NaCl does not behave as two fully independent particles at this concentration [6, 7]. A calculated osmolarity is not a measured osmolality.

B2. Osmolite 1.0 is stated by the manufacturer as 288 mOsm/kg H₂O, with 84.9 g water per 100 mL [8]. Estimate its osmolarity.

Answer

Osmolarity ≈ osmolality × kg water per L = 288 × 0.849 = 244.5 → 245 mOsm/L. This agrees with the label’s 244 within rounding [8]. See enteral formula osmolality.

B3. At what temperature does a 290 mOsm/kg sample freeze (K_f = 1.86 °C·kg/mol)?

Answer

ΔT = 1.86 × 0.290 = 0.539 °C, so the sample freezes at about −0.539 °C [9]. See how osmometers work.

C. Calculated serum osmolality, including the BUN/urea trap

Formula: 2 × Na + Glucose/18 + BUN/2.8 (US), or 2 × Na + Glucose + Urea, all in mmol/L (SI) [10, 11]. Tool: serum osmolality calculator.

C1. Na 140 mmol/L, glucose 90 mg/dL, BUN 14 mg/dL.

Answer

280 + 90/18 + 14/2.8 = 280 + 5.0 + 5.0 = 290 mOsm/kg (calculated). The normal serum range is about 275–295, and ranges vary by laboratory [12].

C2. Na 125, glucose 900 mg/dL, BUN 56 mg/dL.

Answer

250 + 50.0 + 20.0 = 320 mOsm/kg. The sodium is low, but the total osmolality is high because of glucose.

C3. SI units: Na 130, glucose 30 mmol/L, urea 20 mmol/L.

Answer

260 + 30 + 20 = 310 mOsm/kg. In SI units, glucose and urea are added without conversion.

C4. A lab reports urea 30 mg/dL (not BUN). Na 140, glucose 90 mg/dL. Calculate correctly, then show the error from using ÷ 2.8.

Answer
  • Correct: urea 30 ÷ 6.0 = 5.0 mmol/L, so 280 + 5.0 + 5.0 = 290
  • Wrong: 30 ÷ 2.8 = 10.7, giving 280 + 5.0 + 10.7 = 295.7 → 296

The ÷ 2.8 divisor applies to BUN only. Urea mg/dL = BUN mg/dL × 2.14.

C5. Na 140, glucose 99 mg/dL, BUN 14 mg/dL. What is the displayed result?

Answer

280 + 5.5 + 5.0 = 290.5 → 291 (half away from zero). Keep 290.5 unrounded if you go on to compute a gap (see E3).

D. Effective osmolality (tonicity)

Effective osmolality = 2 × Na + Glucose/18. Urea is excluded because it crosses cell membranes freely [13]. Background: isotonic, hypotonic, hypertonic.

D1. Na 140, glucose 90 mg/dL, BUN 14 mg/dL. Calculate the effective osmolality.

Answer

280 + 5.0 = 285 mOsm/kg. BUN does not enter the calculation; changing it would not change the answer.

D2. Na 125, glucose 900 mg/dL. Is the effective osmolality ”> 300”?

Answer

250 + 50.0 = 300. That is not > 300. Threshold statements such as the 2024 consensus HHS criterion “effective osmolality > 300 mOsm/kg” are applied by clinicians with other criteria. They are not a self-check [14].

D3. Na 135, glucose 1,080 mg/dL.

Answer

270 + 60.0 = 330 mOsm/kg (effective). For how high glucose affects the sodium reading, see corrected sodium for hyperglycemia.

E. Osmolar gap, including ethanol

Gap = measured − calculated, using the unrounded calculated value. A gap of 10 or more (Merck ≥ 10) is the commonly used threshold, a convention rather than a sharp boundary [11, 15]. Gaps up to about 15–20 occur in non-toxic states, and > 20 strongly suggests an unmeasured osmole such as a toxic alcohol [16]. Thresholds depend on the equation used, and a normal gap does not exclude toxic alcohol ingestion [17]. Tool: osmolar gap.

E1. C1 inputs, with a measured osmolality of 300 mOsm/kg.

Answer

300 − 290 = 10. That is 10 or more (Merck ≥ 10), so it is flagged as elevated (10–20 tier). It does not identify a cause.

E2. C1 inputs plus ethanol 100 mg/dL. Measured osmolality 350. Calculate the gap with ÷ 3.7 and with ÷ 4.6.

Answer
  • ÷ 3.7 (default; Purssell [18]): calculated = 290 + 27.03 = 317.03 (displays 317). Gap = 350 − 317.03 = 32.97 → 33
  • ÷ 4.6 (alternative): calculated = 290 + 21.74 = 311.74 (displays 312). Gap = 38.26 → 38

Both are in the > 20 tier, which strongly suggests another unmeasured osmole even after ethanol is counted. State which factor you used.

E3. C5 inputs (calculated 290.5), measured 300. What gap is displayed, and what is the common mistake?

Answer

300 − 290.5 = 9.5 → 10, which is flagged. Rounding the calculated value to 291 first gives 9, which is wrong. Compute the gap from the unrounded value.

E4. C1 inputs, measured 287.5. Display the gap.

Answer

287.5 − 290 = −2.5 → −3 (half away from zero). Negative gaps are shown with their sign. JavaScript’s Math.round(-2.5) returns −2, which is wrong.

E5. C1 inputs. Classify a measured osmolality of 310, then one of 311.

Answer
  • 310 − 290 = 20: still the 10–20 “elevated” tier
  • 311 − 290 = 21: the > 20 tier, which strongly suggests an unmeasured osmole such as a toxic alcohol (including ethanol if it was not entered)

Neither number identifies a cause.

F. Tonicity of IV fluids

F1. D5W is about 252 mOsm/L in the bag. Why is it described as hypotonic in the patient?

Answer

Cells take up and metabolize the glucose, which leaves 0 mOsm/L of effective solute. The fluid therefore behaves as free water [19]. Iso-osmolar in the bag does not mean isotonic in the body.

F2. Calculate the osmolarity of 0.45% NaCl (4.5 g/L) and explain “half saline, half free water”.

Answer

4.5 ÷ 58.44 = 77.0 mmol/L, and × 2 = 154 mOsm/L. One litre carries the same Na⁺ and Cl⁻ as 0.5 L of 0.9% NaCl (77 mEq each) plus 0.5 L of electrolyte-free water, so it is hypotonic. (Computed value; no 0.45% NaCl label has been verified for this page.)

G. IV mixing

Mixture osmolarity = total mOsm ÷ total volume. Do not average osmolarities without volume weights. These are arithmetic exercises, not mixing instructions.

G1. 500 mL D5W plus 500 mL 0.9% NaCl.

Answer

252.3 × 0.5 = 126.15 and 308.0 × 0.5 = 154.0. (126.15 + 154.0) ÷ 1.0 L = 280.15 → 280 mOsm/L.

G2. 750 mL D5W plus 250 mL 0.9% NaCl. Compare with a simple average.

Answer

252.3 × 0.75 = 189.2 and 308.0 × 0.25 = 77.0. (189.2 + 77.0) ÷ 1.0 = 266.2 → 266 mOsm/L. The unweighted average (252.3 + 308) ÷ 2 = 280 is wrong for unequal volumes.

G3. D5 + 0.45% NaCl: 50 g/L hydrous dextrose plus 4.5 g/L NaCl.

Answer

252.3 + 154.0 = 406.3 → 406 mOsm/L, which matches the ICU Medical label (406) [20].

H. Parenteral nutrition estimate

Per-litre factors used by Osmolar’s TPN calculator:

  • generic amino acids 10 mOsm/g (product not specified; US labels run 7.88–9.98)
  • hydrous dextrose 5.046 mOsm/g (ICU Medical Dextrose 50% label: 2523 mOsmol/L (calc) ÷ 500 g/L [29]); 20% lipid 1.3 mOsm/g fat [21, 22]
  • NaCl and KCl 2 [24], MgSO₄ 1.0 (Hospira MgSO₄ 50% label: 4.06 mOsmol/mL ÷ 4.06 mEq/mL [30]), sodium phosphate 1.75 mOsm/mEq Na [23]
  • calcium gluconate 1.4667 mOsm/mEq (editorial default; calculated from the B. Braun UK label’s “Theoretical osmolarity: 660 mOsm/l” at 2.25 mmol Ca per 10 mL; other labels give different values, see the calcium gluconate note) [28]

Results are estimates only, and institutional thresholds vary. Tool: TPN osmolarity calculator.

H1. Per litre: AA 30 g, dextrose 50 g, fat 30 g (20% emulsion), Na 50 (NaCl), K 20 (KCl), Mg 5 (sulfate), Ca 5 (gluconate) mEq.

Answer

300 + (50 × 5.046 = 252.3) + 39 + 100 + 40 + 5 + (5 × 1.4667 = 7.3335) = 743.6335 → 744 mOsm/L (estimate). This is below the commonly used 900 mOsm/L peripheral threshold. ASPEN’s recommendation is weak, and thresholds vary by institution [25].

H2. Per litre: AA 35 g, dextrose 100 g, Na 22.5 mEq (NaCl). How is the result classified against 900? What goes wrong with a dextrose factor of 5?

Answer

350 + 504.6 + 45 = 899.6 → 900. Classification uses the displayed integer, so this falls in the ≥ 900 category (Intralipid label wording: “Greater than or equal to 900 mOsm/L must be infused through a central vein” [22]). Classifying on the unrounded 899.6 would wrongly show “900 · below 900”. With dextrose 5 mOsm/g: 350 + 500 + 45 = 895, which is “below”. The dextrose factor alone moves the class. The estimate also carries an error of tens of mOsm [31], so a borderline value calls for pharmacist review.

H3. Repeat H1, but give 20 of the 50 mEq Na as sodium phosphate. What does using the old teaching factor of 3 mOsm/mEq do?

Answer

NaCl 30 × 2 = 60. Na phosphate 20 × 1.75 = 35 (Hospira label: 7 mOsm/mL ÷ 4 mEq/mL) [23]. 300 + 252.3 + 39 + 60 + 35 + 40 + 5 + 7.3335 = 738.6335 → 739 mOsm/L. Using 3 gives 763.6335 → 764, which overstates the phosphate contribution.

I. Oral rehydration solution

Background: ORS vs sports drinks.

I1. WHO reduced-osmolarity ORS contains Na⁺ 75, Cl⁻ 65, K⁺ 20, citrate 10 and glucose 75 mmol/L. What is the total?

Answer

75 + 65 + 20 + 10 + 75 = 245 mOsm/L, below plasma by design [26].

I2. Apple juice is listed at 730 mOsm/L, with sodium 0.4 mmol/L, under “Commonly used beverages (not appropriate for diarrhea treatment)” [27]. What happens if it is diluted 1:1 with water (ideal arithmetic)?

Answer

730 ÷ 2 = 365 mOsm/L, and sodium 0.2 mmol/L. That is still above ORS and plasma, with almost no sodium. Dilution does not make juice an oral rehydration solution. This is arithmetic, not a recipe. Do not make home rehydration mixes for infants or children; ask a clinician or pharmacist.

Educational content, not medical advice. The problems teach arithmetic and are not clinical decisions, fluid prescriptions or mixing instructions. Verify with your institution’s protocols or pharmacist. See our disclaimer and how we check content.

References

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