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Chapter 4: Problem 78

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 is 2.7464 M.

Step by step solution

01

Calculate the mass of glycerol

Using the given density (1.2656 g/mL) and volume (50.000 mL), we can calculate the mass of glycerol using the formula: mass = density × volume Mass of glycerol = (1.2656 g/mL) × (50.000 mL) Mass of glycerol = 63.28 g
02

Calculate the moles of glycerol

We'll now convert the mass of glycerol into moles using the molar mass of glycerol, which is given by the sum of the atomic masses of all atoms in the molecule: Molar mass of glycerol = (3 × M_C) + (8 × M_H) + (3 × M_O) where \(M_C = 12.01 \, g/mol\), \(M_H = 1.008 \, g/mol\), and \(M_O = 16.00 \, g/mol\). Molar mass of glycerol = (3 × 12.01 g/mol) + (8 × 1.008 g/mol) + (3 × 16.00 g/mol) = 92.093 g/mol Now, we can calculate the moles of glycerol using the formula: moles = mass / molar mass Moles of glycerol = (63.28 g) / (92.093 g/mol) = 0.6866 mol
03

Calculate the molarity of the solution

Now we have the moles of glycerol in the solution, and we need to find the molarity. First, we need to convert the solution volume from mL to L: volume = 250.00 mL = 0.25000 L Now use the formula to calculate the molarity: Molarity = moles / volume Molarity of the glycerol solution = (0.6866 mol) / (0.25000 L) = 2.7464 M Thus, the molarity of the glycerol solution is 2.7464 M.

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Most popular questions from this chapter

In each of the following pairs, indicate which has the higher concentration of \(\mathrm{Cl}^{-}\) ion: (a) \(0.10 \mathrm{MAlCl}_{3}\) solution or a $0.25 \mathrm{MLiCl}\( solution, (b) \)150 \mathrm{~mL}\(. of a \)0.05 \mathrm{MMnCl}_{3}\( solution or \)200 \mathrm{~mL}$. of \(0.10 \mathrm{M} \mathrm{KCl}\) solution, (c) a \(2.8 M H C l\) solution or a solution made by dissolving $23.5 \mathrm{~g}\( of \)\mathrm{KCl}\( in water to make \)100 \mathrm{~mL}$ of solution.

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. (a) \(\mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{FeCl}_{3}(a q) \longrightarrow\) (b) $\mathrm{ZnCl}_{2}(a q)+\mathrm{Cs}_{2} \mathrm{CO}_{3}(a q) \longrightarrow$ (c) $\mathrm{Na}_{2} \mathrm{~S}(a q)+\operatorname{CoSO}_{4}(a q) \longrightarrow$

Hard water contains \(\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}\), and \(\mathrm{Fe}^{2+}\), which interfere with the action of soap and leave an insoluble coating on the insides of containers and pipes when heated. Water softeners replace these ions with \(\mathrm{Na}^{+} .\) Keep in mind that charge balance must be maintained. (a) If \(1500 \mathrm{~L}\) of hard water contains \(0.020 \mathrm{M} \mathrm{Ca}^{2+}\) and \(0.0040 \mathrm{M} \mathrm{Mg}^{2+}\), how many moles of Nat are needed to replace these ions? (b) If the sodium is added to the water softener in the form of \(\mathrm{NaCl}\), how many grams of sodium chloride are needed?

Will precipitation occur when the following solutions are mixed? If so, write a balanced chemical equation for the reac- tion. (a) \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\) and \(\mathrm{NaOH},(\mathbf{b}) \mathrm{K}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NH}_{4} \mathrm{NO}_{3}\), (c) \(\mathrm{Na}_{2} \mathrm{~S}\) and \(\mathrm{FeCl}_{3}\)

Using the activity series (Table 4.5 ), write balanced chemical equations for the following reactions. If no reaction occurs, write NR. (a) Nickel metal is added to a solution of copper(II) nitrate, (b) a solution of zinc nitrate is added to a solution of magnesium sulfate, (c) hydrochloric acid is added to gold metal, (d) chromium metal is immersed in an aqueous solution of cobalt(II) chloride, (e) hydrogen gas is bubbled
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