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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

State whether each of the following statements is true or false. Justify your answer in each case. (a) \(\mathrm{NH}_{3}\) contains no \(\mathrm{OH}^{-}\) ions, and yet its aqueous solutions are basic. (b) HF is a strong acid. (c) Although sulfuric acid is a strong electrolyte, an aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) contains more \(\mathrm{HSO}_{4}^{-}\) -ions than \(\mathrm{SO}_{4}^{2-}\) ions.

Using the activity series (Table 4.5), write balanced chemical equations for the following reactions. If no reaction occurs, write NR. (a) Iron metal is added to a solution of copper(II) nitrate, (b) zinc metal is added to a solution of magnesium sulfate, (c) hydrobromic acid is added to tin metal, (d) hydrogen gas is bubbled through an aqueous solution of nickel(II) chloride, (e) aluminum metal is added to a solution of cobalt(II) sulfate.

Complete and balance the following molecular equations, and then write the net ionic equation for each: \begin{equation} \begin{array}{l}{\text { (a) } \operatorname{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow} \\ {\text { (b) } \mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow} \\\ {\text { (c) } \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow}\end{array} \end{equation}

You know that an unlabeled bottle contains an aqueous solution of one of the following: \(A g N O_{3},\) CaCl \(_{2},\) or \(A l_{2}\left(S O_{4}\right)_{3}\) . A friend suggests that you test a portion of the solution with \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and then with NaCl solutions. According to your friend's logic, which of these chemical reactions could occur, thus helping you identify the solution in the bottle? (a) Barium sulfate could precipitate. (b) Silver chloride could precipitate. (c) silver sulfate could precipitate. (d) More than one, but not all, of the reactions described in answers a-c could occur. (e) All three reactions described in answers a-c could occur.

Classify each of the following substances as a nonelectrolyte, weak electrolyte, or strong electrolyte in water: (a) \(\mathrm{H}_{2} \mathrm{SO}_{3}\) , (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (ethanol), \((\mathbf{c}) \mathrm{NH}_{3},(\mathbf{d}) \mathrm{KClO}_{3}\), \((\mathbf{e}) \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free