Problem 61 - Chemistry: An Atoms-Focused Approach

Question

What mass of the following solutions contains 0.100 mol of solute? (a) \(0.334 \mathrm{m} \mathrm{NH}_{4} \mathrm{NO}_{3} ;\) (b) \(1.24 \mathrm{m}\) ethylene glycol, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH} ;(\mathrm{c}) 5.65 \mathrm{m} \mathrm{CaCl}_{2}\)

Step-by-Step Solution

Answer

(a) 0.334 m NH4NO3, (b) 1.24 m ethylene glycol (HOCH2CH2OH), and (c) 5.65 m CaCl2. Answer: The total masses of the solutions are (a) 307.005 g for NH4NO3, (b) 86.807 g for ethylene glycol, and (c) 28.798 g for CaCl2.

Step 1: Recall the formula for molality

Molality (m) is defined as the number of moles of solute per kilogram of solvent. Mathematically, it can be represented as: \(m = \frac{n_{solute}}{m_{solvent}}\) Where \(m\) represents the molality of the solution, \(n_{solute}\) represents the moles of the solute, and \(m_{solvent}\) represents the mass of the solvent in kilograms.

Step 2: Rearrange the formula to solve for the mass of the solvent

We are trying to find the mass of the solvent in each solution, given the moles of solute and the molality. Rearranging the molality formula, we can solve for the mass of the solvent: \(m_{solvent} = \frac{n_{solute}}{m}\)

Step 3: Calculate the mass of the solvent for each solution

Using the values provided in the exercise, we can calculate the mass of the solvent for each solution: (a) 0.334 m NH4NO3 \({m_{solvent}}_{NH4NO3} = \frac{0.100}{0.334} = 0.299 kg\) (b) 1.24 m ethylene glycol (HOCH2CH2OH) \({m_{solvent}}_{ethylene glycol} = \frac{0.100}{1.24} = 0.0806 kg\) (c) 5.65 m CaCl2 \({m_{solvent}}_{CaCl2} = \frac{0.100}{5.65} = 0.0177 kg\)

Step 4: Calculate the mass of solute for each solution

We can use the moles of solute and the molecular weight (M) of each solute to find the mass of solute in each solution: (a) NH4NO3: Molecular weight = 14.01 (N) + 1.01 (H) * 4 + 14.01 (N) + 16.00 (O) * 3 = 80.05 g/mol \(mass_{NH4NO3} = 0.100 mol * 80.05 \frac{g}{mol} = 8.005 g\) (b) Ethylene glycol (HOCH2CH2OH): Molecular weight = 62.07 g/mol \(mass_{ethylene glycol} = 0.100 mol * 62.07 \frac{g}{mol} = 6.207 g\) (c) CaCl2: Molecular weight = 40.08 (Ca) + 35.45 (Cl) * 2 = 110.98 g/mol \(mass_{CaCl2} = 0.100 mol * 110.98 \frac{g}{mol} = 11.098 g\)

Step 5: Calculate the total mass of each solution

Finally, we can find the total mass of each solution by adding the mass of solvent and the mass of solute: (a) \({mass_{solution}}_{NH4NO3} = {m_{solvent}}_{NH4NO3} + mass_{NH4NO3} = 0.299 kg + 8.005 g = 299 g + 8.005 g = 307.005 g\) (b) \({mass_{solution}}_{ethylene glycol} = {m_{solvent}}_{ethylene glycol} + mass_{ethylene glycol} = 0.0806 kg + 6.207 g = 80.6 g + 6.207 g = 86.807 g\) (c) \({mass_{solution}}_{CaCl2} = {m_{solvent}}_{CaCl2} + mass_{CaCl2} = 0.0177 kg + 11.098 g = 17.7 g + 11.098 g = 28.798 g\) Thus, the total mass of each solution containing 0.100 mol of solute is: (a) 307.005 g for NH4NO3, (b) 86.807 g for ethylene glycol, and (c) 28.798 g for CaCl2.

Key Concepts

Moles of Solute

When discussing solutions, it's crucial to understand the term 'moles of solute.' A mole is a basic unit in chemistry that measures the amount of a substance. It is defined as exactly 6.022 x 10\(^{23}\) particles, which could be atoms, molecules, ions, etc. This number is known as Avogadro's number.

In a solution, the solute is the substance being dissolved, while the solvent is the medium in which it is dissolved. Moles of solute are an important measurement because they help us understand the quantity of a solute present in a solution.

• In this exercise, we have 0.100 moles of solute for each different solution.
• This fixed amount allows us to determine the necessary mass of solvent to achieve the specified molality.

Mass of Solvent

The mass of the solvent is a critical value in preparing solutions, especially when determining the molality. Molality is expressed as moles of solute per kilogram of solvent. In this exercise, we were required to calculate the mass of the solvent using the rearranged molality formula: \[ m_{solvent} = \frac{n_{solute}}{m} \]where \(n_{solute}\) is the number of moles of solute and \(m\) is the molality of the solution.

The mass of the solvent gives us an idea of how much of it is required for dissolving a specific number of moles of solute to reach a desired concentration.

• We calculated the solvent mass for different solutes: NH\(_4\)NO\(_3\), ethylene glycol, and CaCl\(_2\).
• This calculation is essential for practical laboratory and industrial applications, ensuring accurate solution preparation and consistency.

Molecular Weight

Molecular weight is a key concept in chemistry that refers to the mass of a single molecule of a chemical compound. It is expressed in atomic mass units (amu) or grams per mole (g/mol). By knowing the molecular weight, we can easily convert between moles and grams. It becomes crucial when you need to calculate the mass from the number of moles.

To calculate molecular weight, you'll need the periodic table and the chemical formula of the substance. Add up the atomic masses of all atoms in the molecule.

• For NH\(_4\)NO\(_3\), the molecular weight was calculated as 80.05 g/mol.
• Ethylene glycol, with a formula of HOCH\(_2\)CH\(_2\)OH, has a molecular weight of 62.07 g/mol.
• CaCl\(_2\) has a molecular weight of 110.98 g/mol.

Understanding molecular weight is essential for determining how much of a substance you need and helps ensure correct proportions in chemical reactions and solution preparations.