Question 9.17: The solution of glucose commonly used intravenously has a co...
The solution of glucose commonly used intravenously has a concentration of 5.0% (w/v) glucose. What is the osmolarity of this solution? The molecular weight of glucose is 180 amu.
ANALYSIS Since glucose is a molecular substance that does not give ions in solution, the osmolarity of the solution is the same as the molarity. Recall from Section 9.7 that a solution of 5.0% (w/v) glucose has a concentration of 5.0 g glucose per 100 mL of solution, which is equivalent to 50 g per liter of solution. Thus, finding the molar concentration of glucose requires a mass to mole conversion.
BALLPARK ESTIMATE One liter of solution contains 50 g of glucose (MW= 180 g/mol). Thus, 50 g of glucose is equal to a little more than 0.25 mol, so a solution concentration of 50 g/L is equal to about 0.25 osmol, or 0.25 M.
STEP 1: Identify known information. We know the (w/v)% concentration of the glucose solution.
5.0\% (w/v) = \frac{5.0 g glucose}{100 mL solution} \times 100\%STEP 2: Identify answer and units. We are looking for osmolarity, which in this case is equal to the molarity of the solution because glucose is a molecular substance and does not dissociate into ions.
Osmolarity = Molarity = ?? mol/liter
STEP 3: Identify conversion factors. The (w/v)% concentration is defined as grams of solute per 100 mL of solution, and molarity is defined as moles of solute per liter of solution. We will need to convert from milliliters to liters and then use molar mass to convert grams of glucose to moles of glucose.
\frac{g glucose}{100 \cancel{ mL}} \times \frac{1000 \cancel{ mL}}{L} = \frac{g glucose}{L} \\ \frac{\cancel{g glucose}}{L} \times \frac{1 mol glucose}{180 \cancel{g glucose} } = \frac{moles glucose}{L}STEP 4: Solve. Starting with the (w/v)% glucose concentration, we first find the number of grams of glucose in 1 L of solution and then convert to moles of glucose per liter.
\left(\frac{ 5.0 g glucose}{100 \cancel{ mL} solution }\right) \left(\frac{1000 \cancel{ mL}}{1 L}\right) = 50 \frac{g glucose}{L solution} \\ \left(\frac{50 \cancel{g} glucose}{1 L}\right) \left(\frac{1 mol}{180 \cancel{g} }\right) = 0.28 M glucose = 0.28 osmolBALLPARK CHECK: The calculated osmolarity is reasonably close to our estimate of 0.25 osmol