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 \(1.50 \mathrm{MHNO}_{3}(a q)\), (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 of the solutions are: (a) \(0.3825\) mol of HNO3, (b) \(1.5\) mol of NaCl, and (c) \(0.00329\) mol of sucrose.

Step 1: Find the number of moles of HNO3 in a 1.50M aqueous solution

To find the number of moles of HNO3, we need to multiply the volume of the solution by its molarity: Number of moles = molarity × volume Given that molarity of HNO3 = 1.50 M and volume of solution = 255 mL. Convert the volume to liters by dividing by 1000: volume = 0.255 L Number of moles = 1.50 M × 0.255 L = 0.3825 mol of HNO3

Step 2: Find the number of moles of NaCl in a 1.50m aqueous solution

To find the number of moles of NaCl, we need to use the molality and given mass of the solution. Molality = moles of solute / kg of solvent Given that molality of NaCl = 1.50m and mass of solution = 50.0 mg. Convert the mass to grams by dividing by 1000: mass = 0.050 g Number of moles = molality × kg of solvent Assume 1 kg of solvent (water) Number of moles = 1.50m × 1 kg = 1.5 mol of NaCl

Step 3: Find the number of moles of sucrose in a 1.50% aqueous solution

To find the number of moles of sucrose, we need to use the mass percent and given mass of the solution. Mass percent of sucrose = (mass of sucrose / mass of solution) × 100 Given that mass percent of sucrose = 1.50% and mass of solution = 75.0 g First, find the mass of sucrose in the solution: \(mass\,of\,sucrose = (mass\,of\,solution × mass\,percent\,of\,sucrose)/100 \) \(mass\,of\,sucrose = (75g × 1.50\%)/100 = 1.125 \, g\) Now, find the moles of sucrose using its molar mass (C12H22O11) = 342 g/mol: Number of moles = mass of sucrose / molar mass Number of moles = 1.125 g / 342 g/mol = 0.00329 mol of sucrose

Key Concepts

Molarity

Molarity is a key concept in chemistry, especially when dealing with solutions. It tells us how many moles of a solute are present in one liter of solution. The formula is:

• Molarity (M) = \( \frac{\text{moles of solute}}{\text{liters of solution}} \)

To find molarity, first ensure your volume is in liters. If given in milliliters, divide by 1000. This is crucial for accurate calculations.
Consider a solution where you have 1.50 M HNO3. If you know the volume of this solution in liters, you can easily calculate moles by multiplying its molarity by this volume. For example:
- Volume (L) = 0.255 L (from 255 mL)- Moles = 1.50 M × 0.255 L = 0.3825 mol of HNO3
This straightforward method makes molarity a versatile tool in chemical calculations.

Molality

Molality is another way to express the concentration of a solution. It measures moles of solute per kilogram of solvent, not per liter of solution. Here's the formula:

• Molality (m) = \( \frac{\text{moles of solute}}{\text{kilograms of solvent}} \)

Molality is particularly useful when dealing with temperature changes since it doesn't rely on volume, which can expand or contract. For example, if you have 50 mg of a solution with a molality of 1.50 m NaCl, you first convert the mass to grams (50 mg = 0.050 g) and assume the mass of the solvent is roughly 1 kg (since the solute's mass is so small):
- Moles = 1.50 m × 1 kg = 1.5 mol of NaCl
This approach gives a consistent measure regardless of temperature fluctuations.

Mass Percent

Mass percent is used to describe how much of a particular component is in a mixture. It's expressed as the mass of the solute divided by the total mass of the solution, multiplied by 100:

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

Imagine you have 75 g of a 1.50% sucrose solution. To find how much sucrose is present, you'd calculate:
- Mass of sucrose = \((75 g \times \frac{1.50}{100}) = 1.125 g\)
Then, use the molar mass of sucrose (C\(_{12}\)H\(_{22}\)O\(_{11}\) = 342 g/mol) to find moles:
- Moles of sucrose = \( \frac{1.125 g}{342 g/mol} = 0.00329 \) mol
Mass percent helps determine the proportion of components, valuable in many chemical and industrial processes.