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Chapter 13: Problem 51

Calculate the number of moles of solute present in each of the following aqueous solutions: (a) 600 \(\mathrm{mL} .\) of 0.250 \(M \operatorname{SrBr}_{2},(\mathbf{b}) 86.4 \mathrm{g}\) of \(0.180 \mathrm{m} \mathrm{KCl},(\mathrm{c}) 124.0 \mathrm{g}\) of a solution that is 6.45\(\%\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) by mass.

Short Answer

Expert verified
The number of moles of solute present in each solution are as follows: (a) 0.15 moles of SrBr鈧�, (b) 0.01553 moles of KCl, and (c) 0.0444 moles of glucose.

Step by step solution

01

a. 600 mL of 0.250 M SrBr鈧� solution

To find the number of moles in this solution, we will use the molarity formula: \(moles = Molarity * Volume\). Given: Molarity, M = 0.250 mol/L Volume, V = 600 mL = \(600 *10^{-3}\) L (Converting milliliters to liters) Now, let's plug the values into the formula: \(moles = (0.250) * (600 *10^{-3})\) Calculating the moles, we get: \(moles = 0.15\) mol There are 0.15 moles of SrBr鈧� in the solution.
02

b. 86.4 g of 0.180 m KCl solution

In this case, we are given the mass (Mass_sol) and molality (m) of a KCl solution. The molality formula is: \(molality = moles(sol) / mass(kg) \). We can rearrange this formula to find the moles of solute as: \(moles(sol) = molality * mass(kg) \). Given: Molality, m = 0.180 mol/kg Mass of the solution, Mass_sol = 86.4 g Now, let's plug the values into the formula: \(moles(KCl) = (0.180) * (86.4 *10^{-3})\) (Converting grams to kg) Calculating the moles, we get: \(moles(KCl) = 0.01553\) mol There are 0.01553 moles of KCl in the solution.
03

c. 124.0 g of a solution that is 6.45% glucose by mass

In this case, we are given the mass of a glucose solution and its mass percent concentration. We can calculate the mass of glucose in the solution by multiplying the mass of the solution by the mass percent. Then we can use the molar mass of glucose to calculate the number of moles. Given: Mass of solution, Mass_sol = 124.0 g Mass percent of glucose = 6.45 % The mass of glucose in the solution: Mass_glucose = \(124 g * \frac{6.45}{100}\) Calculating the mass, we get: Mass_glucose = 7.998 g The molar mass of glucose, \(C_6 H_{12} O_6\), is calculated as: Molar_mass = \(6 * (12.01 g/mol) + 12 * (1.01 g/mol) + 6 * (16.00 g/mol) = 180.18 g/mol\) Now we can use the mass of glucose and the molar mass to find the moles of glucose: \(moles(glucose) = \frac{Mass_{glucose}}{Molar_{mass}}\) Plugging in the values: \(moles(glucose) = \frac{7.998}{180.18}\) Calculating the moles, we get: \(moles(glucose) = 0.0444\) mol There are 0.0444 moles of glucose in the solution.

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Key Concepts

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

Molarity Calculations
Molarity is a common way to express the concentration of a solution, and it depends on both the number of moles of solute and the volume of the solution. For molarity calculations, we use the formula:
  • \[ \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]
To calculate the number of moles of solute from the molarity, we rearrange the formula to: \[ \text{Moles of solute} = \text{Molarity} \times \text{Volume of solution in liters} \]For example, if you have a solution of 600 mL of 0.250 M \(\text{SrBr}_2\), first convert the volume from milliliters to liters (600 mL = 0.600 L), then multiply by the molarity: \[ \text{Moles of } \text{SrBr}_2 = 0.250 \times 0.600 = 0.15 \text{ moles} \]This simple method allows you to quickly find out how much solute is dissolved in a given solution.
Molality Calculations
Molality is another way to express concentration, but unlike molarity, it does not depend on volume. Instead, it is based on the mass of the solvent in kilograms:
  • \[ \text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \]
To find the moles when given molality, use the formula:\[ \text{Moles of solute} = \text{Molality} \times \text{Mass of solvent in kg} \]For instance, with 86.4 g of a 0.180 m KCl solution, convert the mass of the solvent to kilograms (86.4 g = 0.0864 kg). Then multiply by the molality:\[ \text{Moles of } KCl = 0.180 \times 0.0864 = 0.01553 \text{ moles} \]Molality is advantageous in scenarios where temperature changes can affect solution volume, as it relies solely on mass.
Mass Percent Concentration
Mass percent concentration is a way to describe the concentration of a solute in a solution as a percentage of the total mass. It is calculated using the following relation:
  • \[ \text{Mass percent} = \left( \frac{\text{mass of solute}}{\text{total mass of solution}} \right) \times 100 \% \]
To find the number of moles of solute, first determine the mass of the solute using its mass percent, then convert this mass to moles using the solute's molar mass. Take, for example, a solution that is 124 g with a 6.45% glucose by mass. The mass of glucose can be calculated as follows:\[ \text{Mass of glucose} = 124 \times \frac{6.45}{100} = 7.998 \text{ g} \]Then, using the molar mass of glucose (\(C_6H_{12}O_6\)), which is 180.18 g/mol, the moles of glucose can be calculated:\[ \text{Moles of glucose} = \frac{7.998}{180.18} = 0.0444 \text{ moles} \]Mass percent offers a straightforward understanding of a solution's composition, especially suitable for solid-liquid mixtures.

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Most popular questions from this chapter

Proteins can be precipitated out of aqueous solution by the addition of an electrolyte; this process is called "salting out" the protein. (a) Do you think that all proteins would be precipitated out to the same extent by the same concentration of the same electrolyte? (b) If a protein has been salted out, are the protein-protein interactions stronger or weaker than they were before the electrolyte was added? (c) A friend of yours who is taking a biochemistry class says that salting out works because the waters of hydration that surround the protein prefer to surround the electrolyte as the electrolyte is added; therefore, the protein's hydration shell is stripped away, leading to protein precipitation. Another friend of yours in the same biochemistry class says that salting out works because the incoming ions adsorb tightly to the protein, making ion pairs on the protein surface, which end up giving the protein a zero net charge in water and therefore leading to precipitation. Discuss these two hypotheses. What kind of measurements would you need to make to distinguish between these two hypotheses?

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Acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1.80 \(\mathrm{M}\) LiBr solution in acetonitrile is 0.826 \(\mathrm{g} / \mathrm{cm}^{3} .\) Calculate the concentration of the solution in (a) molality, (b) mole fraction of LiBr, (c) mass percentage of \(\mathrm{CH}_{3} \mathrm{CN} .\)

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The solubility of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water is 208 \(\mathrm{g}\) per 100 \(\mathrm{g}\) of water at \(15^{\circ} \mathrm{C}\) . A solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water at \(35^{\circ} \mathrm{C}\) is formed by dissolving 324 \(\mathrm{g}\) in 100 \(\mathrm{g}\) of water. When this solution is slowly cooled to \(15^{\circ} \mathrm{C},\) no precipitate forms. (a) Is the solution that has cooled down to \(15^{\circ}\) Cunsaturated, saturated, or supersaturated? (b) You take a metal spatula and scratch the side of the glass vessel that contains this cooled solution, and crystals start to appear. What has just happened? (c) At equilibrium, what mass of crystals do you expect to form?
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