Question

The total concentration of dissolved particles in blood is \(0.30 \mathrm{M}\). An intravenous (IV) solution must be isotonic with blood, which means it must have the same concentration. (a) To relieve dehydration, a patient is given \(100 . \mathrm{mL} / \mathrm{h}\) of \(\mathrm{IV}\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) for \(2.5 \mathrm{~h}\). What mass (g) of glucose does she receive? (b) If isotonic saline \((\mathrm{NaCl})\) is used, what is the molarity of the solution? (c) If the patient is given \(150 . \mathrm{mL} / \mathrm{h}\) of IV saline for \(1.5 \mathrm{~h}\), how many grams of \(\mathrm{NaCl}\) does she receive?

Short answer

The patient receives 13.5 g of glucose, the molarity of isotonic saline is 0.30 M, and the patient receives 3.945 g of NaCl.

Step-by-step solution

Title - Determining Volume of IV Glucose Received

Calculate the total volume of IV glucose solution administered over the given period. The flow rate is 100 mL/h over 2.5 hours.

Calculation of Total Volume (Glucose)

Total volume = Flow rate × Time = (100 mL/h) × (2.5 h) = 250 mL

Title - Determining Moles of Glucose Received

Use the total concentration of dissolved particles to determine the moles of glucose received. The molarity (M) is 0.30 M and volume (V) is 250 mL (convert to Liters).

Calculation of Moles of Glucose

Moles of glucose = Molarity × Volume (L) = 0.30 M × 0.250 L = 0.075 moles

Title - Calculating Mass of Glucose

Convert the moles of glucose to mass using the molar mass of glucose (C₆H₁₂O₆). The molar mass of glucose is 180 g/mol.

Mass of Glucose Calculation

Mass (g) = Moles × Molar Mass = 0.075 moles × 180 g/mol = 13.5 g

Title - Determining Molarity of Isotonic Saline

Since the IV solution must be isotonic with blood, it must have the same concentration of dissolved particles, 0.30 M.

Isotonic Saline Molarity

The molarity of isotonic saline (NaCl) is 0.30 M.

Title - Determining Volume of IV Saline Received

Calculate the total volume of isotonic saline administered over the given period. The flow rate is 150 mL/h over 1.5 hours.

Calculation of Total Volume (Saline)

Total volume = Flow rate × Time = (150 mL/h) × (1.5 h) = 225 mL

Title - Determining Moles of NaCl Received

Use the total concentration of dissolved particles to determine the moles of NaCl received. The molarity (M) is 0.30 M and volume (V) is 225 mL (convert to Liters).

Calculation of Moles of NaCl

Moles of NaCl = Molarity × Volume (L) = 0.30 M × 0.225 L = 0.0675 moles

Title - Calculating Mass of NaCl

Convert the moles of NaCl to mass using the molar mass of NaCl. The molar mass of NaCl is 58.44 g/mol.

Mass of NaCl Calculation

Mass (g) = Moles × Molar Mass = 0.0675 moles × 58.44 g/mol = 3.945 g

Key concepts

IV solutions

IV solutions are crucial in medical treatments. They provide patients with necessary fluids, electrolytes, and medications directly into their veins. This method ensures quick absorption and efficient delivery.
In this exercise, we specifically look at glucose and saline IV solutions used in treating dehydration.
The goal is to make these solutions isotonic, meaning they have the same concentration as blood to avoid cell damage.

molarity calculations

Molarity is essential for understanding concentrations in solutions. It refers to the number of moles of solute per liter of solution. To calculate molarity, you can use the formula:
Molarity (M) = Moles of solute / Volume of solution (L)
For example, if you have a blood concentration of 0.30 M, an isotonic IV solution must also be 0.30 M to match it. This ensures that the IV solution will not cause cells to shrink (hypertonic) or swell (hypotonic).

mass calculation of solutes

To find the mass of a solute in a solution, you first need the number of moles of the solute. Once you have that, use the molar mass to get the mass. The formula is:
Mass (g) = Moles × Molar Mass (g/mol)
For instance, in step 6 of the solution, we found the patient received 0.075 moles of glucose. Knowing glucose has a molar mass of 180 g/mol, we calculate:
0.075 moles × 180 g/mol = 13.5 g
This method applies to any solute, such as NaCl in isotonic saline.

isotonic saline

Isotonic saline is a common IV solution used in healthcare. It contains sodium chloride (NaCl) at a concentration that matches the osmolarity of blood. This avoids any negative effects on cells.
The exercise tells us that the molarity for an isotonic solution must be 0.30 M, the same as the blood concentration. If we administer saline at this molarity, it ensures cell stability, providing a safe and effective hydration method.

glucose in IV therapy

Glucose in IV therapy helps treat dehydration by providing not just fluids but also energy. When the body is dehydrated, it might lack blood glucose.
In this exercise, the IV glucose solution has to be isotonic, with the same total concentration as blood (0.30 M). Over 2.5 hours, the patient receives 250 mL of IV glucose, translating to 13.5 g of glucose.
This careful calculation ensures the patient gets both hydration and necessary sugars without disturbing their cellular balance.