Question

How would you prepare the following aqueous solutions? (a) \(3.10 \times 10^{2} \mathrm{~g}\) of \(0.125 \mathrm{~m}\) ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) from ethylene glycol and water (b) \(1.20 \mathrm{~kg}\) of 2.20 mass \(\% \mathrm{HNO}_{3}\) from 52.0 mass \(\% \mathrm{HNO}_{3}\)

Short answer

Solution (a): Use 274.5 g of ethylene glycol and water to make 310 g of solution. Solution (b): Mix 50.77 g of 52.0% HNO₃ with 1.149 kg of water.

Step-by-step solution

Understand the Problem

We need to prepare two different kinds of aqueous solutions. One involves a molal (m) solution of ethylene glycol, and the other involves a mass percentage solution of nitric acid.

Calculate the amount of ethylene glycol for Solution (a)

Molality (m) is defined as moles of solute per kilogram of solvent. Given that the solution has a molality of 0.125 m and mass of the solution is 310 g (0.310 kg), we will use the formula to find the amount of ethylene glycol needed: \[ \text{molality} = \frac{\text{moles of solute}}{\text{kg of solvent}} \] Let the mass of ethylene glycol be \( m_{\text{eg}} \) and the molar mass of ethylene glycol \( \text{C}_2\text{H}_6\text{O}_2 \) be 62.07 g/mol. Thus, we have: \[ 0.125 = \frac{\frac{m_{\text{eg}}}{62.07}}{0.310 - m_{\text{eg}}} \]

Solve for the mass of ethylene glycol (Solution (a))

Rearrange the formula and solve for \( m_{\text{eg}} \): \[ 0.125 = \frac{m_{\text{eg}}}{62.07 \times (0.310 - m_{\text{eg}})} \]Multiply both sides by \( 62.07 \times (0.310 - m_{\text{eg}}) \):\[ 0.125 \times 62.07 \times (0.310 - m_{\text{eg}}) = m_{\text{eg}} \]Expand and simplify:\[ 7.75875 \times 0.310 - 7.75875 \times m_{\text{eg}} = m_{\text{eg}} \]\[ 2.4052075 - 7.75875 m_{\text{eg}} = m_{\text{eg}} \]\[ 2.4052075 = 8.75875 m_{\text{eg}} \]Divide both sides by 8.75875:\[ m_{\text{eg}} \thickapprox 0.2745 \text{ kg or 274.5 g} \]

Prepare Solution (a)

To prepare 310 g of a 0.125 m ethylene glycol solution, measure 274.5 g of ethylene glycol and add enough water to bring the total mass to 310 g.

Calculate the amount of 52.0% HNO₃ for Solution (b)

Use the given mass percentage formula by mixing the concentrated 52.0% HNO₃ solution to make 1.20 kg of a 2.20% HNO₃ solution. The mass of HNO₃ desired is: \[ 1.20 \text{ kg} \times 0.022 = 0.0264 \text{ kg of HNO₃} \]

Determine the mass of 52.0% HNO₃ solution needed

Using the mass percentage of the concentrated solution, \[ \frac{0.264 \text{ kg}}{0.52} \thickapprox 0.050769 \text{ kg or 50.77 g of 52.0% HNO₃ solution} \]

Calculate the amount of water needed (Solution (b))

Subtract the mass of the concentrated solution from the total mass of the final solution:\[ 1.20 \text{ kg} - 0.050769 \text{ kg} = 1.149231 \text{ kg of water} \]

Prepare Solution (b)

Measure 50.77 g of the 52.0% HNO₃ solution and add enough water to achieve a total mass of 1.20 kg (which will require approximately 1.149 kg of water).

Key concepts

Molality

Molality (m) is a way to express the concentration of a solution. Unlike molarity, which is based on the volume of solution, molality depends on the mass of the solvent. It's defined as the number of moles of solute per kilogram of solvent. The formula for calculating molality is: \( \text{molality} = \frac{\text{moles of solute}}{\text{kg of solvent}} \). This unit is particularly useful when dealing with temperature variations, as it does not change with temperature. To find the molality of a solution, you need to know the mass of the solvent and the number of moles of the solute.

Mass Percentage

Mass percentage is a way to express the concentration of a solution by indicating the mass of the solute present in a certain mass of the solution. It's calculated by using the formula: \( \text{mass percentage} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100 \). This measurement is very useful for understanding the proportion of solute in a mixture. For instance, if you have a solution with 2.20% nitric acid (HNO₃), it means that 2.20 grams of HNO₃ are present in 100 grams of the solution. Mass percentage is especially useful for mixing different concentrations to prepare a new solution.

Solution Preparation

Preparing a solution involves accurately measuring and mixing chemicals. To prepare a specific concentration, you need to understand the relation between solute and solvent. For example, if you are preparing a 0.125 m solution of ethylene glycol (C₂H₆O₂), you calculate the moles of ethylene glycol needed and adjust the mass of water accordingly to achieve the desired molality. Accurately measuring masses and volumes ensures that the resultant solution has the correct concentration. Always ensure to mix well after adding solute to the solvent. This not only ensures uniform distribution but also avoids any concentration gradients that can affect the solution's properties.

Ethylene Glycol

Ethylene glycol (C₂H₆O₂) is commonly used in antifreeze and as a coolant. It is a colorless, odorless, and relatively viscous liquid. In aqueous solution preparation, knowing the molar mass of ethylene glycol (62.07 g/mol) is crucial for calculating the amount needed. For instance, when preparing a 0.125 m solution with 310 grams of total mass, you need approximately 274.5 grams of ethylene glycol. Understanding the properties and correct handling of ethylene glycol is essential for safe and accurate solution preparation.

Nitric Acid

Nitric acid (HNO₃) is a highly corrosive and toxic strong acid used mainly for fertilizers, explosives, and in chemical synthesis. When preparing a dilute nitric acid solution from a concentrated one, precision is crucial. For example, to prepare 1.20 kg of a 2.20% HNO₃ solution from a 52.0% stock solution, you compute the mass of HNO₃ required and then determine how much of the 52.0% solution is needed to provide this mass. Subsequently, water is added to achieve the desired final mass. Safety is paramount when handling nitric acid due to its corrosive nature.