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Glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3},\) is a substance used extensively in the manufacture of cosmetic s, 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

Expert verified
The molarity of the glycerol solution made by dissolving 50.000 mL glycerol at \(15^{\circ} C\) in enough water to make 250.00 mL of solution is 2.748 M.

Step by step solution

01

Calculate the mass of glycerol

Given the density and volume of glycerol, we can determine its mass using the following formula: Mass of glycerol = Density x Volume Density of glycerol = 1.2656 g/mL Volume of glycerol = 50.000 mL Mass of glycerol = 1.2656 g/mL x 50.000 mL Mass of glycerol = 63.280 g
02

Calculate the molar mass of glycerol

To find the molar mass of glycerol, \({C}_{3} {H}_{8} {O}_{3}\), we will sum the molar masses of all its constituent elements. Molar mass of glycerol = (3 x Molar mass of Carbon) + (8 x Molar mass of Hydrogen) + (3 x Molar mass of Oxygen) Given: Molar mass of Carbon (C) = 12.01 g/mol Molar mass of Hydrogen (H) = 1.008 g/mol Molar mass of Oxygen (O) = 16.00 g/mol Molar mass of glycerol = (3 x 12.01) + (8 x 1.008) + (3 x 16.00) Molar mass of glycerol = 36.03 + 8.064 + 48.00 Molar mass of glycerol = 92.094 g/mol
03

Calculate the molarity of the glycerol solution

Now, we have the mass of glycerol, its molar mass, and the volume of the solution. To find the molarity, we can use the formula: Molarity = (Mass of solute / Molar mass of solute) / Volume of solution (in Liters) We know: Mass of glycerol (solute) = 63.280 g Molar mass of glycerol = 92.094 g/mol Volume of the solution = 250.00 mL = 0.25000 L Now we can plug the values into the formula: Molarity = (63.280 g / 92.094 g/mol) / 0.25000 L Molarity = 0.6870 mol/0.25000 L Molarity = 2.748 M The molarity of the glycerol solution is 2.748 M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Glycerol
Glycerol is a simple polyol compound, which means it's a type of alcohol with multiple hydroxyl (OH) groups. Its chemical formula is \( \mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3} \). Glycerol is well-known for its sweet taste and is a key ingredient in many industries. It is often used in cosmetics, such as skin lotions and creams, because it helps to retain moisture and give a smooth texture. In the food industry, it's used as a sweetener and a food preservative. Additionally, glycerol is utilized in the production of antifreeze and plastics because of its ability to lower the freezing point. Importantly, glycerol is a water-soluble liquid which makes it versatile for different applications, especially when making solutions.
The Process of Mass Calculation
Mass calculation is a crucial step in chemistry for determining how much of a substance you have. To find the mass of a compound like glycerol, when you know its density and volume, use the formula:
  • Mass = Density \( \times \) Volume
For glycerol, which has a density of 1.2656 g/mL and a volume of 50.000 mL, multiply these two values:
  • Mass of glycerol = 1.2656 g/mL \( \times \) 50.000 mL = 63.280 g
This equation shows a direct relationship: increasing the volume of the liquid, at constant density, increases the mass. This relationship helps scientists and students accurately determine the amount needed for reactions or solutions.
The Significance of Molar Mass
Molar mass is a fundamental concept in chemistry that links the mass of a molecule to its amount in moles. It represents the mass of one mole of a substance. The molar mass is calculated by adding together the atomic masses of all atoms in the molecule's chemical formula. For glycerol (\( \mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3} \)), the calculation involves:
  • Carbon (C): 3 atoms \( \times \) 12.01 g/mol = 36.03 g/mol
  • Hydrogen (H): 8 atoms \( \times \) 1.008 g/mol = 8.064 g/mol
  • Oxygen (O): 3 atoms \( \times \) 16.00 g/mol = 48.00 g/mol
Summing these gives the molar mass of glycerol:
  • Molar mass of glycerol = 92.094 g/mol
This value is essential when converting between grams and moles, facilitating various calculations including molarity.
Learning About Density Calculation
Density is defined as the mass per unit volume of a substance. It's a key property that helps determine how much mass is contained in a given volume. The formula for density is:
  • Density = \( \frac{\text{Mass}}{\text{Volume}} \)
For glycerol, knowing its density (1.2656 g/mL) allows you to perform calculations to find the mass of a specific volume, like in the problem we're working on. Conversely, if you have a mass and need to find volume, this property still supports those computations. Practically, density helps in many areas of science and engineering as it affects fluid flow, buoyancy, and material properties. Understanding density and its computations can provide insights into how substances will behave in different conditions and solutions.

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Most popular questions from this chapter

(a) How many milliliters of 0.120 \(\mathrm{M}\) HCl are needed to completely neutralize 50.0 \(\mathrm{mL}\) of 0.101 \(\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution? (b) How many milliliters of 0.125 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize 0.200 \(\mathrm{g}\) of NaOH? \((\mathrm{c})\) If 55.8 \(\mathrm{mL}\) of a BaCl \(_{2}\) solution is needed to precipitate all the sulfate ion in a 752 -mg sample of \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) what is the molarity of the BaCl\(_{2}\) solution?

Glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3},\) is a substance used extensively in the manufacture of cosmetic s, 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.

(a) You have a stock solution of 14.8 \(\mathrm{M} \mathrm{NH}_{3}\) . How many milliliters of this solution should you dilute to make 1000.0 \(\mathrm{mL}\) of 0.250 \(\mathrm{MNH}_{3} ?(\mathbf{b})\) you take a 10.0 -mL portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{L},\) what will be the concentration of the final solution?

The distinctive odor of vinegar is due to aceticacid, \(\mathrm{CH}_{3} \mathrm{COOH},\) which reacts with sodium hydroxide according to: $$\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NaOH}(a q) \longrightarrow_{\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCH}_{3} \mathrm{COO}(a q)}$$ If 3.45 \(\mathrm{mL}\) of vinegar needs 42.5 \(\mathrm{mL}\) of 0.115 \(\mathrm{M} \mathrm{NaOH}\) to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00 -qt sample of this vinegar?

Write the balanced molecular and net ionic equations for each of the following neutralization reactions: \begin{equation} \begin{array}{l}{\text { (a) Aqueous acetic acid is neutralized by aqueous barium }} \\ {\text { hydroxide. }} \\ {\text { (b) Solid chromium(II) hydroxide reacts with nitrous acid. }} \\ {\text { (c) Aqueous nitric acid and aqueous ammonia react. }}\end{array} \end{equation}

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