Question

Glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3},\) is a substance used extensively in the manufacture of cosmetics, foodstuffs, antifreeze, and plastics. Glycerol is a water-soluble liquid with a density of \(1.2656 \mathrm{~g} / \mathrm{mL}\) at \(15^{\circ} \mathrm{C}\). Calculate the molarity of a solution of glycerol made by dissolving \(50.000 \mathrm{~mL}\) glycerol at \(15^{\circ} \mathrm{C}\) in enough water to make \(250.00 \mathrm{~mL}\) of solution.

Short answer

The molarity of the glycerol solution is calculated as follows: First, find the mass of glycerol by multiplying the density and volume: \(mass = (1.2656 \mathrm{~g/mL})(50.000 \mathrm{~mL})\). Next, convert the mass to moles using the molar mass of glycerol, \(92.09 \mathrm{~g/mol}\): \(moles = mass ÷ molar \ mass\). Finally, calculate the molarity by dividing the number of moles by the total solution volume in liters (\(0.25000 \mathrm{~L}\)): \(Molarity = (moles)/(0.25000 \mathrm{~L})\).

Step-by-step solution

Calculate mass of glycerol

Given, the density of glycerol is 1.2656 g/mL, and the volume of dissolved glycerol is 50.000 mL. To find the mass of glycerol, use the formula: mass = density × volume. So, mass of glycerol = (1.2656 g/mL)(50.000 mL)

Convert mass to moles

To convert the mass of glycerol to moles, divide the mass by the molar mass of glycerol (which is the sum of the molar masses of its constituent elements, i.e., \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\)). The molar mass of glycerol is \(3(12.01)+8(1.01)+3(16.00) = 92.09 \mathrm{~g/mol}\). Now, the number of moles of glycerol = mass ÷ molar mass.

Calculate the molarity

The molarity is calculated by dividing the number of moles by the total solution volume in liters. The total volume of the solution is given as 250.00 mL, which is equal to 0.25000 L. Then, calculate the molarity as: Molarity = (number of moles)/(total solution volume in L) Put together the previous steps to find the molarity of the glycerol solution.

Key concepts

Density of Liquids

Understanding the density of a liquid is key to calculating its mass from a given volume. Density is a property that describes the mass of a substance per unit volume. For liquids, it is usually measured in grams per milliliter (g/mL). Knowing the density allows you to convert a liquid volume into a mass.
For glycerol, given a density of 1.2656 g/mL, this means each milliliter of glycerol weighs 1.2656 grams. To find the mass of glycerol when you have 50 mL, multiply the density by the volume:

• Mass of glycerol = 1.2656 g/mL × 50.000 mL
• This results in a mass of 63.28 g of glycerol.

This mass is essential for further calculations, like determining the number of moles.

Molar Mass

The molar mass of a compound tells you how much one mole of that substance weighs in grams. To calculate the molar mass, summing up the atomic masses of all the atoms in a molecule is necessary. Let's break this down for glycerol \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\).

• Carbon (C): 3 atoms × 12.01 g/mol = 36.03 g/mol
• Hydrogen (H): 8 atoms × 1.01 g/mol = 8.08 g/mol
• Oxygen (O): 3 atoms × 16.00 g/mol = 48.00 g/mol

Adding these together, you get the molar mass of glycerol:

• \(Molar mass = 36.03 + 8.08 + 48.00 = 92.11 \, \text{g/mol}\)

Using this, you can convert the mass of glycerol to moles by dividing the mass by the molar mass.

Solution Volume Conversion

When preparing solutions, converting the volume to the appropriate units is crucial for accurate calculation of molarity. Molarity is defined as the number of moles of solute per liter of solution. In many problems, solution volume is given in milliliters, so conversion to liters is necessary.
For example, if the total solution volume is given as 250.00 mL, convert this to liters by using the conversion factor (1 L = 1000 mL):

• Total solution volume in liters = 250.00 mL ÷ 1000 = 0.25000 L

Once in liters, you can compute molarity using:

• Molarity = \(\frac{\text{Number of moles}}{\text{Total solution volume in liters}}\)
• Plug in the values obtained from prior calculations to find the molarity of your solution.

This ensures consistency and accuracy in your calculations.