Problem 51 - Chemistry The Central Science

Question

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.

Step-by-Step Solution

Answer

The number of moles of solute present in each solution are as follows: (a) 0.15 moles of SrBr₂, (b) 0.01553 moles of KCl, and (c) 0.0444 moles of glucose.

Step 1: a. 600 mL of 0.250 M SrBr₂ 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 SrBr₂ in the solution.

Step 2: 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.

Step 3: 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.

Key Concepts

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.