Problem 52 - Chemistry: The Central Science in SI Units, Global Edition

Question

Calculate the number of moles of solute present in each of the following solutions: (a) \(255 \mathrm{~mL}\) of \(1.50 \mathrm{M} \mathrm{HNO}_{3}(a q)\) (b) \(50.0 \mathrm{mg}\) of an aqueous solution that is \(1.50 \mathrm{~m} \mathrm{NaCl}\), (c) \(75.0 \mathrm{~g}\) of an aqueous solution that is \(1.50 \%\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) by mass.

Step-by-Step Solution

Answer

The number of moles of solute present in each solution are: (a) 0.3825 mol of HNO3 (b) 3.96 × 10^{-4} mol of NaCl (c) 3.29 × 10^{-3} mol of sucrose

Step 1: Solve for problem (a):

To find the moles of HNO3, we are given its molarity and the volume of the solution. We can use the formula: Moles of solute = Molarity × Volume (in L) We are given Molarity=1.50 M and Volume= 255 mL. First, convert the volume to liters: Volume (L) = 255 mL × (1 L / 1000 mL) = 0.255 L Now, calculate the moles of HNO3: Moles of HNO3 = (1.50 M) × (0.255 L) = 0.3825 mol

Step 2: Solve for problem (b):

To find the moles of NaCl, we are given the mass of the solution and its molality. We can use the formula: Moles of solute = (Molality × Mass of solvent)/(Molar mass of solute) We are given Molality = 1.50 m, and the Mass of the solution = 50.0 mg and the molar mass of NaCl = 58.44 g/mol. First, convert the mass of the solution to grams: Mass (g) = 50.0 mg × (1 g / 1000 mg) = 0.0500 g Now, calculate the mass of the solvent: Mass of solvent = Mass of solution - Mass of solute Since the molality is given, we can write the equation: Mass of Solute = Molality × Mass of solvent First, we need to find the mass of solvent. Rearranging, Mass of solvent = Mass of Solute / Molality We can denote the mass of solvent by x and the mass of solute by y: 0.0500 g = x + y y = (1.50 m) × x y = (1.50) × (0.0500 g - y) Solving for y, we get y = 0.023148 g Now, find the moles of NaCl: Moles of NaCl = mass of solute / Molar mass of solute Moles of NaCl = 0.023148 g / 58.44 g/mol = 3.96 × 10^{-4} mol

Step 3: Solve for problem (c):

To find the moles of sucrose, we are given the mass percentage, and we can first find mass of the solute and then we can find moles using the molar mass. We are given mass percentage = 1.50 % and the Mass of the solution = 75.0 g and the molar mass of sucrose = 342.3 g/mol. First, calculate the mass of sucrose: Mass of sucrose = (1.50 %) × (75.0 g) = 1.125 g Now, find the moles of sucrose: Moles of sucrose = mass of solute / Molar mass of solute Moles of sucrose = 1.125 g / 342.3 g/mol = 3.29 × 10^{-3} mol Now we have the number of moles of solute present in each solution: (a) 0.3825 mol of HNO3 (b) 3.96 × 10^{-4} mol of NaCl (c) 3.29 × 10^{-3} mol of sucrose

Key Concepts

Molarity

Molarity is an essential concept in chemistry, as it defines the concentration of a solute in a solution. It tells us how many moles of a solute are present per liter of solution. The formula to calculate molarity (\(M\)) is:

• Molarity (\(M\)) = Moles of solute / Volume of solution in liters

This calculation helps chemists understand how concentrated a solution is, which is crucial for predicting how substances will react together.

In practice, you may be given the volume of the solution in milliliters, as in our example, and you need to convert it to liters before using it in the molarity formula. For instance, \(255 \text{ mL}\) becomes \(0.255 \text{ L}\).

Knowing the molarity, you can easily calculate the number of moles in the solution by multiplying the molarity by the volume (converted to liters). This is what we did to find there are \(0.3825 \text{ mol}\) of \(\text{HNO}_3\) in our solution. Understanding molarity allows you to work with solutions effectively, making it a vital tool in chemistry.

Molality

Molality is another way to express the concentration of a solution. It is particularly useful when dealing with temperature changes because it is based on mass, not volume, which can expand or contract. Unlike molarity, molality is defined as the moles of solute per kilogram of solvent. The formula for molality (\(m\)) is:

• Molality (\(m\)) = Moles of solute / Mass of solvent in kilograms

This indicator is helpful in assessing properties that depend on the total amount of solute like boiling point elevation or freezing point depression.

For example, to find the molality of a given solution, you need to determine how much solvent you have, as seen when we rearranged the solution's total mass to isolate the solvent's mass. Then, you divide the number of moles of solute by the mass of the solvent in kilograms. This yielded us a molality that allowed us to find there are \(3.96 \times 10^{-4} \text{ mol}\) of \(\text{NaCl}\) in the solution. The dependency on mass rather than volume makes molality an important concentration measure in environments with fluctuating temperatures.

Mass Percentage

Mass percentage is a straightforward concept used to express how much of the total solution is made up by the solute. It is calculated by taking the mass of the solute divided by the total mass of the solution and multiplying by 100 to get a percentage. The formula is:

• Mass Percentage = (Mass of solute / Mass of solution) \(\times 100\)

This form of concentration expression is very intuitive because it directly tells you what portion of the solution is solute. It is particularly useful when preparing or analyzing solutions where you need to know the exact composition by mass.

In our example, we calculated the mass percentage to see what part of a given 75 g solution was sucrose. From that, we obtained the mass of sucrose (the solute), allowing us to determine there are \(3.29 \times 10^{-3} \text{ mol}\) of sucrose present. This concept is incredibly useful in industries dealing with mass and in everyday contexts like cooking or material preparation where mass proportions are critical.