Question

Calculate the number of moles of solute present in each of the following solutions: (a) \(185 \mathrm{~mL}\) of \(1.50 \mathrm{M}\) \(\mathrm{HNO}_{3}(a q)\), (b) \(50.0 \mathrm{mg}\) of an aqueous solution that is \(1.25 \mathrm{~m} \mathrm{NaCl}\), (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.

Short answer

In summary, the number of moles of solute present in each solution is: (a) \(0.2775\ mol\) of \(\mathrm{HNO_3}\), (b) \(0.0625\ mol\) of \(\mathrm{NaCl}\), and (c) \(0.00329\ mol\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\).

Step-by-step solution

Part (a) - Using Molarity Formula

Molarity (M) is defined as moles of solute per liter of solution. The given volume is in milliliters, so it needs to be converted to liters. After that, we can use the formula for molarity to determine the number of moles of \(\mathrm{HNO}_3\). Molarity = Moles of solute / Volume of solution We are given: - Volume of solution: 185 ml or \(0.185 L\) - Molarity: \(1.50 M\) Moles of solute = Volume of solution × Molarity Moles of \(\mathrm{HNO_3}\) = \(0.185 L × 1.50 M\) Moles of \(\mathrm{HNO_3}\) = \(0.2775 mol\)

Part (b) - Using Molality Formula

Molality (m) is defined as moles of solute per kilogram of solvent. We are given the mass of the solution, so we need to find the mass of the solvent. After that, we can use the formula for molality to determine the number of moles of \(\mathrm{NaCl}\). Molality = Moles of solute / Mass of solvent (in kg) We are given: - Mass of solution: \(50.0 mg\) or \(0.050 g\) - Molality: \(1.25 m\) Since molality is given, we need to find the mass of the solvent (water). We know that molality is the ratio of moles of solute to the mass of solvent in kg: \(1.25 m = \frac{moles\ of\ NaCl}{0.050 kg}\) Now, we solve for moles of solute: Moles of \(\mathrm{NaCl} = 1.25 m × 0.050 kg = 0.0625 mol\)

Part (c) - Using Mass Percent Formula

Mass percent is defined as the mass of the solute in the solution divided by the total mass of the solution, multiplied by 100. We are given: - Mass of solution: \(75.0 g\) - Mass percent of sucrose: \(1.50\% \) First, find the mass of sucrose in the solution: \(1.50\% = \frac{mass\ of\ sucrose}{75.0 g}\) Mass of sucrose = \(75.0 g × 0.015\) Mass of sucrose = \(1.125 g\) Now, we will find the molar mass of sucrose (\(\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}\)). Molar mass of sucrose = \(12 × (12.01 g/mol) + 22 × (1.01 g/mol) + 11 × (16.00 g/mol)\) Molar mass of sucrose = \(342.3 g/mol\) Finally, we will find the number of moles of sucrose. Moles of sucrose = \(\frac{mass\ of\ sucrose}{molar\ mass\ of\ sucrose}\) Moles of sucrose = \(\frac{1.125 g}{342.3 g/mol}\) Moles of sucrose = \(0.00329 mol\)

Key concepts

Understanding Molarity

Molarity is a way of expressing the concentration of a solution. It tells us how many moles of solute are present in one liter of solution. Think of it as measuring the strength of a lemonade by how much lemon juice (solute) is in a liter of lemonade (solution).
To calculate molarity, you use the formula:

• Molarity (M) = Moles of solute / Liters of solution

In our problem, we convert the volume from milliliters to liters because molarity uses the liter as the unit of volume. So, 185 mL becomes 0.185 L.
Using the given molarity of 1.50 M, we find the moles of solute (HNO3) by multiplying the volume in liters by the molarity:

• Moles of HNO3 = 0.185 L × 1.50 M = 0.2775 mol

This means there are 0.2775 moles of nitric acid in the solution.

Exploring Molality

Molality is another way to measure concentration, closely related to molarity, but with a key difference. Instead of liters of solution, it measures moles of solute per kilogram of solvent. This unit stays constant because mass doesn’t change with temperature, unlike volume.
To compute molality, the formula used is:

• Molality (m) = Moles of solute / Kilograms of solvent

In our exercise, the mass of the solution is given in milligrams, which we convert to grams ( 50.0 mg = 0.050 g ) and then to kilograms to align with the formula ( 0.050 kg ).
Given the molality (1.25 m), we find the moles of NaCl by multiplying:

• Moles of NaCl = 1.25 m × 0.050 kg = 0.0625 mol

Thus, the solution contains 0.0625 moles of sodium chloride.

Decoding Mass Percent

Mass percent tells us what portion of the total mass is made up by one component, expressed as a percentage. It's a bit like saying 10% of your body is water. For chemical solutions, it shows how much of the solute is in the total solution mass.
The mass percent formula is:

• Mass percent = (Mass of solute / Mass of solution) × 100

In our problem, the solution's mass is given as 75.0 g and the sucrose is 1.50% of this. First, find the actual mass of sucrose:

• Mass of sucrose = 75.0 g × 1.50% = 1.125 g

After finding the mass of sucrose, we determine its moles using the molar mass of sucrose ( 342.3 g/mol ):

• Moles of sucrose = 1.125 g / 342.3 g/mol = 0.00329 mol

This shows that 0.00329 moles of sucrose are present in the solution.