Question

A typical IV used in hospitals is dextrose 5$\mathrm{\%}$ in water (called D5W). This solution is injected into veins through an IV to replace lost fluids and to provide carbohydrates. Injectable medicines are also delivered to the body using the D5W IV. D5W contains 5.0 g dextrose monohydrate $({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\cdot {\mathrm{H}}_2\mathrm{O})$ per 100.0 $\mathrm{mL}$ of solution. Assuming a density of 1.01 $\mathrm{g}/{\mathrm{cm}}^3$ , calculate the molarity and molality of D5W.

Short answer

The molarity of the D5W solution is $0.252\text{mol/L}$, and its molality is $0.262\text{mol/kg}$.

Step-by-step solution

Calculate the molar mass of dextrose monohydrate

From the molecular formula (C6H12O6·H2O), we obtain the molar mass as follows: Molar mass of dextrose monohydrate = 6 × (mass of C) + 12 × (mass of H) + 6 × (mass of O) + 2 × (mass of H) + 1 × (mass of O) Considering mass of C = 12.01 g/mol, mass of H = 1.01 g/mol, and mass of O = 16.00 g/mol. Plugging in the values we get: Molar mass = 6 × 12.01 + 12 × 1.01 + 6 × 16.00 + 2 × 1.01 + 1 × 16.00 Molar mass = 72.06 + 12.12 + 96.00 + 2.02 + 16.00 Molar mass = 198.2 g/mol

Calculate the molarity of D5W

In order to calculate the molarity of the D5W solution, we first need to find out how many moles of dextrose monohydrate are present in the solution. With the mass of dextrose monohydrate (5.0 g) given and the molar mass (198.2 g/mol) calculated, we can calculate the moles as follows: Moles of dextrose monohydrate = mass / molar mass Moles of dextrose monohydrate = 5.0 g / 198.2 g/mol Moles of dextrose monohydrate = 0.0252 mol Now, we can find molarity using the formula: Molarity (M) = moles of solute / volume of solution (in liters) First, convert the volume of the solution from mL to L: 100.0 mL = 0.1 L Now calculate the molarity: Molarity (M) = 0.0252 mol / 0.1 L Molarity (M) = 0.252 mol/L

Calculate the molality of D5W

In order to find the molality, we need to find the mass of the solvent (water) in the solution. We already know the mass of dextrose monohydrate in the solution (5.0 g) and we can find the total mass of the solution using its density. Given density = 1.01 g/cm³ and volume = 100 mL, let's calculate the mass of the solution: Total mass of the solution = density × volume Total mass of the solution = 1.01 g/cm³ × 100 cm³ (recall 1 mL = 1 cm³) Total mass of the solution = 101 g Now, we can find the mass of the solvent (water): Mass of water = total mass of the solution - mass of dextrose monohydrate Mass of water = 101 g - 5.0 g Mass of water = 96 g Now, we can find molality using the formula: Molality (m) = moles of solute / mass of solvent (in kg) First, convert the mass of the solvent from g to kg: 96 g = 0.096 kg Now calculate the molality: Molality (m) = 0.0252 mol / 0.096 kg Molality (m) = 0.262 mol/kg Thus, the molarity of the D5W solution is 0.252 mol/L, and its molality is 0.262 mol/kg.

Key concepts

D5W Solution

D5W stands for Dextrose 5% in Water. This solution is commonly used in hospitals as an intravenous (IV) fluid. It's crucial for replenishing fluids and providing a source of carbohydrates. The 5% indicates that there are 5 grams of dextrose per 100 milliliters of water. This precise concentration ensures the solution is isotonic, meaning it has a similar osmotic pressure as human blood, which makes it safe for intravenous administration. D5W is also used to deliver other injectable medicines. Understanding the composition and properties of D5W is vital for calculating concentrations like molarity and molality.

Dextrose Monohydrate

Dextrose monohydrate refers to the sugar, specifically glucose, that is bound with water. Its molecular formula is ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\cdot {\mathrm{H}}_2\mathrm{O}$. This formula indicates that each molecule of dextrose carries an additional water molecule. When considering its molar mass, you account for both the glucose unit and the water. By multiplying the atomic masses (C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol), you find the molar mass of dextrose monohydrate to be 198.2 g/mol. This value is crucial for computing how many moles are present in a given mass.

Density and Concentration

Density is defined as mass per unit volume. In the context of D5W, the density is given as 1.01 g/cm³. This means that for every cubic centimeter, the solution weighs 1.01 grams. To find the total mass of a D5W solution, multiply density by volume. For example, 101 grams corresponds to 100 mL, since 1 mL equals 1 cm³. Understanding density helps determine the mass of the entire solution, which is necessary when calculating molality. Concentration, on the other hand, refers to how much solute (in this case, dextrose) is present in the solution. It includes measurements like molarity and molality, describing the ratio of solute to solution or solvent.

Moles Calculation

To convert mass into moles, use the formula: moles = mass/molar mass. This conversion becomes essential when determining concentration measures like molarity and molality. For example, in the D5W solution, with a known mass of dextrose monohydrate (5.0 g) and its calculated molar mass (198.2 g/mol), the moles come out to be 0.0252 mol. Knowing the moles of dextrose allows you to calculate molarity, which is moles per liter of solution, and molality, which is moles per kilogram of solvent. These calculations are fundamental in chemistry for preparing solutions with a specific concentration.