Question

D5W refers to one of the solutions used as an intravenous fluid. It is a \(5 \%\) by mass solution of dextrose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in water. The density of \(\mathrm{D} 5 \mathrm{~W}\) is \(1.08 \mathrm{~g} / \mathrm{ml}\). The molarity of the solution is (a) \(0.3 \mathrm{M}\) (b) \(0.6 \mathrm{M}\) (c) \(0.28 \mathrm{M}\) (d) \(0.26 \mathrm{M}\)

Short answer

0.3 M

Step-by-step solution

Understanding the Percent Solution

D5W is a 5% by mass solution of dextrose, which means for every 100 grams of solution, there are 5 grams of dextrose.

Calculating the Mass of Dextrose in 1 Liter of Solution

To find the mass of dextrose in 1 liter of D5W, we need to calculate the mass of the solution first. Since the density is 1.08 g/ml, 1 liter (or 1000 ml) of solution has a mass of 1000 ml * 1.08 g/ml = 1080 grams. Now, 5% of 1080 grams is the mass of dextrose: 0.05 * 1080 g = 54 grams.

Converting Grams of Dextrose to Moles

The molar mass of dextrose \(C_{6}H_{12}O_{6}\) is calculated from its formula: (6 * 12.01) + (12 * 1.01) + (6 * 16.00) = 72.06 + 12.12 + 96.00 = 180.18 g/mol. To find the number of moles of dextrose, we divide the mass of dextrose by the molar mass: 54 g / 180.18 g/mol.

Calculating the Molarity of the Solution

Now that we have the moles of dextrose, we can calculate the molarity, which is moles of solute per liter of solution. From Step 3 we know the moles of dextrose in 1 liter, we divide the number of grams by the molar mass of dextrose to find moles. Finally, since the density is given for 1 liter, we can directly use the moles for molarity calculation.

Key concepts

Understanding Percent Composition

Percent composition provides a way to describe the concentration of a component in a mixture or solution. In the context of the D5W solution, percent composition refers to the ratio of the mass of the solute (dextrose) to the mass of the whole solution (dextrose plus water) expressed as a percentage.

To calculate it, we simply take the mass of the solute and divide it by the total mass of the solution, and then multiply the result by 100. For example, a 5% by mass solution of dextrose means that for every 100 grams of D5W, there are 5 grams of dextrose.

Importance of Density in Calculations

Density is a crucial physical property of a substance that represents its mass per unit volume, typically expressed in grams per milliliter (g/ml) or kilograms per cubic meter (kg/m³). When it comes to solutions like D5W, knowing the density allows us to calculate the mass of the solution if the volume is known.

Applying Density in Molarity Calculations

Using density, we can convert from volume to mass, which is the first step in determining the amount of solute present. From there, we can carry out further calculations to find moles and, eventually, molarity, which is a measure of solution concentration. In the case of D5W with a density of 1.08 g/ml, this measure is vital to find the mass of the dextrose in a certain volume of the solution.

Molar Mass and Its Role in Molarity

Molar mass is the mass of one mole of a substance and it is a bridge between the mass of a substance and the number of moles. We calculate it by summing the atomic masses of all the atoms in a molecule of the substance, typically expressed in grams per mole (g/mol).

From Molar Mass to Moles

In calculations of molarity, which is the number of moles of solute per liter of solution, knowing the molar mass allows us to convert the mass of solute (in grams) into moles. This step is critical because molarity depends on moles, not grams. For dextrose (C_{6}H_{12}O_{6}), with each carbon, hydrogen, and oxygen contributing to the total molar mass, knowing this value is fundamental to determining how many moles of dextrose are present in a liter of D5W solution.

Determining Solution Concentration: Molarity

Molarity, denoted by M, is a measurement of the concentration of a solution. It indicates the number of moles of solute dissolved in one liter of solution. It's a very useful concentration metric in chemistry because it allows for straightforward calculations in reactions that involve solutions.

To calculate molarity, we need to know two things: the number of moles of the solute and the volume of the solution in liters. Using the previous steps, where we found the mass of dextrose in the solution, calculated its molar mass, and converted the mass to moles, we can directly calculate the molarity by dividing the moles of solute by the volume of the solution in liters. This gives us the molarity of the D5W solution, which showcases the connection between the molar mass of a compound and the overall concentration of a solution.