Problem 52 - Chemistry The Central Science

Question

Calculate the number of moles of solute present in each of the following solutions: (a) \(255 \mathrm{~mL}\) of \(1.50 \mathrm{M} \mathrm{HNO}_{3}(a q),(\mathbf{b})\) \(50.0 \mathrm{mg}\) 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

In solution (a), there are 0.3825 moles of HNO₃ present. In solution (b), there are 0.075 moles of NaCl present. In solution (c), there are 0.00329 moles of sucrose present.

Step 1: Solution (a)

For the first part, we have been given the volume of the solution and its molarity (concentration). Molarity is defined as moles of solute per liter of solution. We can use the formula: moles = Molarity × Volume (in liters). First, we need to convert the volume from milliliters (mL) to liters (L) by dividing by 1000. \( V_{L} = \frac{255}{1000} \mathrm{L} = 0.255 \mathrm{L} \) Now, we can calculate the moles of solute: \( \mathrm{moles} = \left(1.50 \frac{\mathrm{moles}}{\mathrm{L}}\right)(0.255 \mathrm{L}) = 0.3825 \mathrm{moles} \) So, in solution (a), there are 0.3825 moles of HNO₃ present.

Step 2: Solution (b)

In the second part, we are given the mass of the solution and its molality (concentration). Molality is defined as moles of solute per kilogram of solvent. We can use the formula: moles = Molality × Mass of solvent (in kg). First, we need to convert the mass from milligrams (mg) to kilograms (kg) by multiplying by 0.001. \( m_{\mathrm{kg}} = 50.0\cdot 0.001 \mathrm{kg} = 0.05 \mathrm{kg} \) Now, we can calculate the moles of solute: \( \mathrm{moles} = \left(1.50 \frac{\mathrm{moles}}{\mathrm{kg}}\right)(0.05 \mathrm{kg}) = 0.075 \mathrm{moles} \) So, in solution (b), there are 0.075 moles of NaCl present.

Step 3: Solution (c)

For the third part, we are given the mass of the solution and the mass percent of sucrose present. Mass percent (%) is defined as the mass of solute divided by the mass of the solution, multiplied by 100. First, we need to find the mass of the solute. We can use the formula: Mass of solute = (Mass percent × Mass of solution) / 100 Mass of sucrose = (1.50% × 75.0 g) / 100 = 1.125 g To calculate the moles of sucrose present, we need to convert the mass of sucrose to moles using the molar mass of sucrose, which is: \( \mathrm{M_{Sucrose}} = 12(\mathrm{C}) + 22(\mathrm{H}) + 11(\mathrm{O}) \) \( \mathrm{M_{Sucrose}} = (12\times 12.01)+(22\times 1.01)+(11\times 16) \mathrm{g~mol}^{-1} = 342.3 \mathrm{g~mol}^{-1} \) Using the mass of sucrose and its molar mass, we can calculate the moles of sucrose: \( \mathrm{Moles} = \frac{\mathrm{Mass}}{\mathrm{Molar~mass}} \) \( \mathrm{Moles} = \frac{1.125\,\mathrm{g}}{342.3\,\mathrm{g~mol^{-1}}} = 0.00329~\mathrm{moles} \) So, in solution (c), there are 0.00329 moles of sucrose present.

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute divided by the volume of the solution in liters. This unit helps us understand how much of a substance is dissolved in a given amount of liquid. For instance, if you have a molarity of 1.50 M (moles per liter) of \( ext{HNO}_3\) in 255 milliliters of solution, you can calculate the number of moles by multiplying the molarity by the volume in liters. First, convert 255 mL to liters by dividing by 1000, giving 0.255 L. Then, simply use the formula:

• Moles = Molarity \( \times \) Volume (in L)

So, the moles of \( ext{HNO}_3\) present are calculated as: \( ext{Moles} = 1.50 \times 0.255 = 0.3825\) moles. This shows how molarity assists in determining the quantity of solute in a solution, which is essential in lab preparations and chemical reactions.

Molality

Molality is another concentration term frequently used in chemistry, particularly when reactions involve considerable temperature changes. It considers the number of moles of solute per kilogram of solvent, independent of the solution's overall volume.To find the molality, you convert the mass of the solvent from milligrams to kilograms by multiplying by 0.001. For example, if you have 50 mg of solvent, it converts to 0.05 kg. With a given molality, say 1.50 m, you can find the number of moles of solute using:

• Moles = Molality \( \times \) Mass of solvent (in kg)

Applying this to the given solution, you get: \( ext{Moles} = 1.50 \times 0.05 = 0.075\) moles. This calculation helps to determine how a solution will behave under varying physical conditions, such as elevation in boiling points or depression in freezing points.

Mass Percent

Mass percent, or weight percent, is a simple way to express the concentration of a component in a mixture or solution. For instance, in a solution, it is given by the mass of solute divided by the total mass of the solution, multiplied by 100 to get a percentage. You can apply this concept by first calculating the mass of the solute using the formula:

• Mass of solute = \( \frac{\text{Mass percent} \times \text{Mass of solution}}{100}\)

For a solution with 1.50% sucrose in 75 grams of solution, this gives: \( ext{Mass of sucrose} = \frac{1.50 \times 75}{100} = 1.125\) grams. Once the mass of sucrose is known, you can easily find the number of moles by dividing the mass of sucrose by its molar mass, showing how mass percent can simplify complex mixture analyses.

Molar Mass

The molar mass is a fundamental concept that relates the mass of a substance to the number of moles. It is the mass of one mole of a chemical compound and is expressed in grams per mole (g/mol). An important application of molar mass is converting between grams and moles, allowing for calculated stoichiometry in reactions.To calculate the molar mass of sucrose (C\(_{12}\)H\(_{22}\)O\(_{11}\)), sum up the average atomic masses of all atoms present in the compound:

• Carbon (C): 12 atoms \( \times 12.01\, \text{g/mol}\)
• Hydrogen (H): 22 atoms \( \times 1.01\, \text{g/mol}\)
• Oxygen (O): 11 atoms \( \times 16\, \text{g/mol}\)

Adding these gives a molar mass of 342.3 g/mol.Using this, if you have a certain mass of sucrose, you can calculate how many moles that represents by dividing the mass by the molar mass, making it an indispensable tool in chemical calculations and practical experiments.