Lesson
Count Particles and Name the Unit
Distinguish solute particles, electrical charge, concentration and total amount.
- Particle counts
- Ions versus formula units
- Osmolarity and osmolality
- Supplied molecular weight
A molecule remains one particle
One calcium plus two chloride
Charge +2 is not two particles
Start with the requested unit
The book defines osmolarity as osmotic particle amount per liter, commonly mOsmol/L. A total milliosmole amount is mOsmol without a volume denominator. Osmolality uses a kilogram basis, mOsmol/kg. Name the requested unit before selecting the equation; do not replace a total particle answer with a per-liter concentration. The book relates particle number to osmotic pressure and distinguishes the temperature dependence of volume from a kilogram basis. It notes that osmolarity and osmolality can have similar numerical values in dilute solutions; their units remain distinct.
Use the book particle table
In the supplied calculation model, dextrose and mannitol each count as one particle. KCl, NaCl, sodium acetate and MgSO4 each count as two. CaCl2 counts as three; sodium citrate, Na3C6H5O7, counts as four. This follows the formula-unit dissociation used in the book exercises. These counts are the model inputs for the original exercises here.
One specified ion is one particle
A problem that supplies sodium ions already specifies the dissociated species, so use one particle per ion. The same applies to a specified calcium ion: its +2 charge does not make it two particles. Distinguish one Ca2+ ion from a CaCl2 formula unit, which supplies one calcium and two chloride ions in the book model.
Match the molecular weight to the stated form
Use the molecular weight supplied for the stated compound form. The source uses 198 g/mol for one dextrose exercise and 180 g/mol when it specifies anhydrous dextrose. Do not silently substitute a remembered value. The book also permits calculating formula-unit molecular weight from supplied atomic values; one Na plus one Cl gives 23 + 35.5 = 58.5 g/mol.
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Lesson
Calculate Osmolarity per Liter
Convert strength to grams per liter before applying the molecular weight and particle count.
- Grams per liter
- Molecular weight
- Particle multiplier
- Rounding
1.5 g/100 mL = 15 g/L
Book calculation model
Nearest whole number
Build the per-liter equation
For the book model, osmolarity in mOsmol/L = [concentration in g/L divided by molecular weight in g/mol] x particle count x 1,000. The ratio first gives mol/L; multiplying by particles gives Osmol/L and by 1,000 gives mOsmol/L. The molecular weight belongs in the denominator.
Convert a percentage before dividing
An original exercise states 1.5% w/v NaCl, MW 58.5 g/mol, and two particles per formula unit. Since 1.5 g/100 mL = 15 g/L, osmolarity = (15/58.5) x 2 x 1,000 = 512.8205... mOsmol/L, or 513 to the nearest whole number. Using 1.5 as though it were already g/L causes a tenfold error.
Handle mg/L and nonionic substances
Another original exercise gives 1,170 mg/L NaCl. Convert to 1.17 g/L, then (1.17/58.5) x 2 x 1,000 = 40 mOsmol/L. A nonionic solute still contributes one particle per molecule: 9.1 g/L mannitol at MW 182 gives 50 mOsmol/L, not zero.
Retain precision until the requested answer
For an original 3% w/v dextrose exercise with supplied MW 198 g/mol and particle count one, 30/198 x 1,000 = 151.5151... mOsmol/L. To the nearest tenth, report 151.5 mOsmol/L. Convert strength and units first, retain intermediate precision, then apply the requested final rounding.
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Lesson
Calculate Total Milliosmoles
Use the mass actually present or multiply osmolarity by the actual volume in liters.
- Total grams
- Actual volume
- Milliosmoles
- Concentration versus amount
MW 111; particles 3
4.44 / 111 x 3 x 1,000
Final volume: 200 mL
Use total mass for total particles
Total mOsmol = [total solute grams divided by MW in g/mol] x particle count x 1,000. An original exercise contains 4.44 g CaCl2 in 200 mL, MW 111 and particle count three. Total amount = 4.44/111 x 3 x 1,000 = 120 mOsmol. The corresponding concentration is 120/0.200 = 600 mOsmol/L; these are different requested quantities.
Multiply a concentration by actual liters
A 250 mL portion at 120 mOsmol/L contains 120 x 0.250 = 30 mOsmol. Conversely, 75 mOsmol in 500 mL gives 75/0.500 = 150 mOsmol/L. Convert mL to L before multiplying or dividing. A one-liter example can give the same numerical value for amount and concentration while retaining different units.
Convert ion mass and count the stated species
An original problem specifies 460 mg sodium ions in 400 mL, MW 23 g/mol. Total mass is 0.460 g and the specified ion is one particle: 0.460/23 x 1,000 = 20 mOsmol. For 160 mg specified calcium ions, MW 40, the total is 0.160/40 x 1,000 = 4 mOsmol. Valence is not used as the particle multiplier in either specified-ion calculation.
Do not normalize an amount answer unnecessarily
If an original exercise specifies 18 g anhydrous dextrose, MW 180 and one particle, its total amount is 100 mOsmol regardless of the stated container volume. Volume becomes necessary if concentration is requested. With final volume 300 mL, the concentration would be 100/0.300 = 333.3333... mOsmol/L, or 333.3 to the nearest tenth.
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Lesson
Combine Components and Solve for Mass
Put every component on the same final-volume basis and reverse the particle equation when mass is requested.
- Component contributions
- Shared final volume
- Reverse equation
- mEq is not particle count
9.1 g / 182 x 1 x 1,000
2.925 g / 58.5 x 2 x 1,000
150 mOsmol in final 0.500 L
Add particle amounts then divide once
An original mixture has final volume 0.500 L, 9.1 g mannitol at MW 182 with one particle, and 2.925 g NaCl at MW 58.5 with two particles. Their contributions are 50 and 100 mOsmol. Total amount is 150 mOsmol; divide by the stated final volume to obtain 300 mOsmol/L. Adding 50 and 100 and labeling the result per liter would omit the final-volume conversion.
Use the same final-volume basis
The source adds dextrose, saline and KCl contributions for the same final one-liter example. Original component values may be added directly in mOsmol/L only when each is expressed for the same finished solution. If 100 and 80 mOsmol/L are specified for a final 1.0 L and 12 mmol KCl is present, the book particle model adds 24 mOsmol/L from KCl, giving 204 mOsmol/L total.
Reverse the equation for required grams
Required mass in grams = target mOsmol/L x final volume in L x MW in g/mol divided by [particle count x 1,000]. An original KCl exercise specifies 200 mOsmol/L in final 300 mL, MW 74.5, and two particles. Required total is 60 mOsmol; mass = 60 x 74.5/2,000 = 2.235 g, or 2.24 g to the nearest hundredth. A reverse check with the unrounded mass returns 200 mOsmol/L.
Separate equivalents from particles
The source converts monovalent KCl equivalents to compound amount, then counts both resulting particles. In an original exercise, 10 mEq KCl is explicitly supplied as 10 mmol KCl in a final 500 mL. Ten mmol formula units x two particles = 20 mOsmol; 20/0.500 = 40 mOsmol/L. Do not use an electrical-equivalent number as though it were automatically the total particle amount. These arithmetic totals alone do not establish compatibility, stability, route or clinical suitability.
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Module practice
Check the connections.
Each attempt draws 10 questions from the complete 25 question bank.
Each attempt draws a fresh set and rearranges the answer choices.
References
Book source and calculation scope.
Original lessons and exercises apply the supplied book calculation model and stated molecular weights. Clinical serum-osmolality assessment and isotonicity adjustment are separate topics.
- RxPrep 2023 Course Book
Chapter 1, required-formulas checklist, printed page 7 (PDF page 19); Chapter 10, Osmolarity, printed pages 139-142 (PDF pages 147-150), ending before Isotonicity.