NAPLEX Question of the Week: Moles, Milliosmoles, and More

Quite possibly the most difficult question of the week we have ever posted. Are you up to the challenge?
NAPLEX Question of the Week: Moles, Milliosmoles, and More

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CA is a 57-year-old male who presents to the emergency room with the following vital signs: T 102.1°F, BP 81/54 mmHg, and RR 30 breaths/min. The diagnosis is presumed sepsis. He is fluid resuscitated with 3 L of Lactated Ringer’s, an aqueous solution containing sodium chloride (molar mass = 58.44 g/mol), sodium lactate (112.07 g/mol), potassium chloride (74.55 g/mol), and calcium chloride (110.98 g/mol). The concentration of sodium chloride is 102.7 mmol/L. Each 100 mL contains 5.5 mOsmol of sodium lactate and 30 mg of potassium chloride. The total osmolarity of the solution is 273 mOsmol/L. How many grams of calcium chloride does the patient receive from the Lactated Ringer’s bolus? Round the final answer to the nearest tenth.




Answer with rationale:


The correct answer is 0.5 g of calcium chloride.


One question format that you will see on the NAPLEX is constructed response, which is basically fill in the blank. This is a great place to incorporate calculation questions. Several paths can be taken to solve the above question, but all involve conversions between mass, concentration, and osmolarity.


Recall that a mole of a substance is 6.02  1023 molecules of that substance. A substance’s molar mass is the mass of one mole of that substance and can be used to convert between mass and moles. Molarity is a measure of concentration in units of moles of substance per liter of solution. Osmolarity is a measure of solute particles per liter of solution. The conversion between molarity and osmolarity depends upon how many particles a substance releases when it is put into solution. A salt like sodium chloride dissociates into two particles (Na+ and Cl-) in solution, so one mole of sodium chloride would be two osmoles of sodium chloride. A sugar like lactose will not dissociate in solution, so one mole of lactose would be one osmole of lactose.


The first step in this problem is to find the osmolarity of the individual substances. To find the osmolarity of sodium chloride, convert molarity to osmolarity. Remember that sodium chloride dissociates into two particles in solution, so the osmolarity will be double the molarity.

 (102.7 mmol NaCl)/(1 L solution)  ×  (2 mOsmol NaCl)/(1 mmol NaCl) = 205.4 mOsmol/L NaCl


The number of osmoles of sodium lactate in 100 mL is given, so find the number of osmoles of sodium lactate in 1 L by converting units.

 (5.5 mOsmol NaC3H5O3)/(100 mL solution)  ×  (1,000 mL)/(1 L) = 55 mOsmol/L NaC3H5O3


The problem also gives the mass of potassium chloride in 100 mL of solution. To find the osmolarity of potassium chloride, convert from mass to moles and then moles to osmoles. Like sodium chloride, potassium chloride dissociates into two particles (K+ and Cl-) in solution.

 (30 mg KCl)/(100 mL solution)  ×  (1,000 mL)/(1 L)  ×  (1 g)/(1,000 mg)  ×  (1 mol KCl)/(74.55 g KCl)  ×  (1,000 mmol)/(1 mol)  ×  (2 mOsmol KCl)/(1 mmol KCl)  = 8.05 mOsmol/L KCl

The osmolarity of calcium chloride is the portion of the total solution osmolarity that is not accounted for by the other three substances.

 273 mOsmol/L - 205.4 mOsmol/L NaCl - 55 mOsmol/L NaC3H5O- 8.05 mOsmol/L KCl=4.55 mOsmol/L CaCl2


The osmolarity of calcium chloride can be converted to concentration and finally mass. The tricky part here is remembering that calcium chloride dissociates into three particles (Ca2+ and 2Cl-) in solution because calcium is a divalent cation. This makes the osmolarity of calcium chloride three times its molarity.

 (4.55 mOsmol CaCl2)/(1 L solution)  ×  (1 mmol CaCl2)/(3 mOsmol CaCl2 )  ×  (1 mol)/(1,000 mmol)  ×  (110.98 g CaCl2)/(1 mol CaCl2 )  × 3 L = 0.5 g CaCl2


Overall, keeping track of your units in a problem like this is just as important as understanding the mathematical process itself. You may have to deal with grams, milligrams, and micrograms all in the same problem. Miscalculating by a factor of 1,000 can have serious consequences in practice.

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