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Calculate the number of moles of solute present in each of the following aqueous solutions: (a) \(750 \mathrm{~mL}\) of \(0.120 \mathrm{M}\) \(\operatorname{SrBr}_{2},(\mathbf{b}) 70.0 \mathrm{~g}\) of \(0.200 \mathrm{~m} \mathrm{KCl},(\mathbf{c}) 150.0 \mathrm{~g}\) of a solution that is \(5.75 \%\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) by mass.

Short Answer

Expert verified
In summary, we have the following number of moles of solute in each solution: (a) \(0.090\) moles of \(\mathrm{SrBr_2}\) in \(750 \mathrm{~mL}\) of \(0.120 \mathrm{M}\) solution. (b) \(0.0112\) moles of \(\mathrm{KCl}\) in \(70.0 \mathrm{~g}\) of \(0.200 \mathrm{~m}\) solution. (c) \(0.0479\) moles of glucose (\(\mathrm{C_6H_{12}O_6}\)) in \(150.0 \mathrm{~g}\) of a solution that is \(5.75\%\) glucose by mass.

Step by step solution

01

Part (a): Calculate moles of SrBr2

To find the number of moles, use the formula: Number of moles = Molarity × Volume (in Liters) Molarity (M) of SrBr2 = 0.120 M Volume (V) of the solution = 750 mL = 0.750 L Number of moles of SrBr2 = 0.120 M × 0.750 L = 0.090 moles
02

Part (b): Calculate moles of KCl

To find the number of moles, use the formula: Number of moles = Molality × Mass of Solvent (in kg) Molality (m) of KCl = 0.200 m Mass of the solvent = 70.0 g - (70.0 g × 0.200) = 56.0 g = 0.056 kg Number of moles of KCl = 0.200 m × 0.056 kg = 0.0112 moles
03

Part (c): Calculate moles of glucose

To find the number of moles, use the formula: Number of moles = Mass of Solute (g) / Molar mass of Solute (g/mol) Molar mass of glucose = 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol Percentage of glucose = 5.75% Total mass of solution = 150.0 g Mass of glucose in the solution = 150.0 g × 0.0575 = 8.625 g Number of moles of glucose = 8.625 g / 180.18 g/mol = 0.0479 moles

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Key Concepts

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

Molarity
Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. The formula to calculate molarity is:\[ \text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]Molarity is a useful concentration unit in laboratory settings because it relates directly to the strength of the solution. When we say a solution is 0.120 M, as in the first example, it means that there are 0.120 moles of the solute (SrBr2) in every liter of that solution. When calculating moles using molarity, always remember to convert the volume from milliliters to liters by dividing by 1000.
Molality
Molality is another way to measure the concentration of a solute in a solution but differs significantly from molarity. While molarity depends on the volume of the entire solution, molality is concerned with the mass of the solvent. The formula for molality is:\[ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kilograms}} \]One of the main advantages of molality is that it does not change with temperature, as it is based on the mass of the solvent rather than volume, which can expand or contract with temperature changes. In the second example, the base calculation uses a mass solvent approach. This shows why molality is especially useful in thermodynamic calculations where temperature dependency could affect the result.
Glucose Percentage
The percentage composition, such as glucose percentage, gives the ratio of the mass of solute to the total mass of the solution, multiplying by 100 to convert it into a percentage. For a 5.75% glucose solution, this means:\[ \text{Mass of glucose} = \text{Total mass of solution} \times \frac{\text{Percentage of glucose}}{100} \]This calculation tells us how much of a particular solute is present relative to the entire solution. In this case, it helps us to determine the actual mass of glucose present in 150.0 g of solution by considering only a smaller portion of the solution's mass as that of glucose. This is useful in dietary contexts and for formulations in lab environments.
Molar Mass
Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a given substance. It is usually expressed in grams per mole (g/mol). Each element has a defined atomic mass, and the molar mass of a compound is calculated by summing the atomic masses of all atoms present in the formula of the compound.For example, glucose (\(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\)) has a molar mass calculated as follows:\[ \text{Molar Mass of } \mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6} = 6(12.01) + 12(1.01) + 6(16.00) = 180.18 \, \text{g/mol} \]Understanding molar mass is crucial for converting between the mass of a substance and the number of moles since the relationship allows chemists to translate real-world measurements into meaningful data for reactions and processes.

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Most popular questions from this chapter

The Henry's law constant for hydrogen gas \(\left(\mathrm{H}_{2}\right)\) in water at \(25^{\circ} \mathrm{C}\) is \(7.7 \times 10^{-6} \mathrm{M} / \mathrm{kPa}\) and the constant for argon (Ar) at \(25^{\circ} \mathrm{C}\) is \(1.4 \times 10^{-5} \mathrm{M} / \mathrm{kPa}\). If the two gases are each present at \(253 \mathrm{kPa}\) pressure, calculate the solubility of each gas.

Indicate the type of solute-solvent interaction (Section 11.2) that should be most important in each of the following solutions: \((\mathbf{a}) \mathrm{CCl}_{4}\) in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right),(\mathbf{b})\) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in water, \((\mathbf{c}) \mathrm{KBr}\) in water, (d) HCl in acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right).\)

What is the molarity of each of the following solutions: (a) \(15.0 \mathrm{~g}\) of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in \(0.250 \mathrm{~mL}\) solution, (b) \(5.25 \mathrm{~g}\) of \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) in \(175 \mathrm{~mL}\) of solution, (c) \(35.0 \mathrm{~mL}\) of \(9.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) diluted to \(0.500 \mathrm{~L} ?\)

At ordinary body temperature \(\left(37^{\circ} \mathrm{C}\right),\) the solubility of \(\mathrm{N}_{2}\) in water at ordinary atmospheric pressure is \(0.015 \mathrm{~g} / \mathrm{L}\) Air is approximately \(78 \mathrm{~mol} \% \mathrm{~N}_{2}\). (a) Calculate the number of moles of \(\mathrm{N}_{2}\) dissolved per liter of blood, assuming blood is a simple aqueous solution. (b) At a depth of \(30.5 \mathrm{~m}\) in water, the external pressure is \(405 \mathrm{kPa}\). What is the solubility of \(\mathrm{N}_{2}\) from air in blood at this pressure? (c) If a scuba diver suddenly surfaces from this depth, how many milliliters of \(\mathrm{N}_{2}\) gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

The presence of the radioactive gas radon \((\mathrm{Rn})\) in well water presents a possible health hazard in parts of the United States. (a) Assuming that the solubility of radon in water with 15.2 kPa pressure of the gas over the water at \(30^{\circ} \mathrm{C}\) is \(0.109 \mathrm{M},\) what is the Henry's law constant for radon in water at this temperature? (b) A sample consisting of various gases contains 4.5 -ppm radon (mole fraction). This gas at a total pressure of 5.07 MPa is shaken with water at \(30^{\circ} \mathrm{C} .\) Calculate the molar concentration of radon in the water.

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