Problem 62 - Chemistry: An Atoms-Focused Approach

Question

How many moles of solute are there in the following solutions? a. \(0.150 \mathrm{m}\) glucose solution made by dissolving the glucose in \(100.0 \mathrm{kg}\) of water b. \(0.028 m \mathrm{Na}_{2} \mathrm{CrO}_{4}\) solution made by dissolving the \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) in \(1000.0 \mathrm{g}\) of water c. \(0.100 \mathrm{m}\) urea solution made by dissolving the urea in \(500.0 \mathrm{g}\) of water

Step-by-Step Solution

Answer

Question: Calculate the number of moles of solute in the following solutions: a. 100.0 kg of a 0.150 m glucose solution. b. 1000.0 g of a 0.028 m Na2CrO4 solution. c. 500.0 g of a 0.100 m urea solution. Answer: a. 15.0 moles of glucose b. 0.028 moles of Na2CrO4 c. 0.05 moles of urea

Step 1: a. Moles of glucose in a \(0.150 m\) glucose solution with \(100.0 kg\) of water

We can calculate the moles of glucose using the given molality and the mass of water: $$moles\ of\ glucose = molality \times mass\ of\ water$$ $$moles\ of\ glucose = 0.150\ \mathrm{m} \times 100.0\ \mathrm{kg}$$ $$moles\ of\ glucose = 15.0$$ Answer: There are 15.0 moles of glucose in the solution.

Step 2: b. Moles of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) in a \(0.028 m\) \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) solution with \(1000.0 g\) of water

We can calculate the moles of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) using the given molality and the mass of water. First, we need to convert the mass of water to kilograms: $$1000.0\ \mathrm{g} = 1.0\ \mathrm{kg}$$ Then, we can calculate the moles of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) using the formula: $$moles\ of\ \mathrm{Na}_{2} \mathrm{CrO}_{4} = molality \times mass\ of\ water$$ $$moles\ of\ \mathrm{Na}_{2} \mathrm{CrO}_{4} = 0.028\ \mathrm{m} \times 1.0\ \mathrm{kg}$$ $$moles\ of\ \mathrm{Na}_{2} \mathrm{CrO}_{4} = 0.028$$ Answer: There are 0.028 moles of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) in the solution.

Step 3: c. Moles of urea in a \(0.100 m\) urea solution with \(500.0 g\) of water

We can calculate the moles of urea using the given molality and the mass of water. First, we need to convert the mass of water to kilograms: $$500.0\ \mathrm{g} = 0.500\ \mathrm{kg}$$ Then, we can calculate the moles of urea using the formula: $$moles\ of\ urea = molality \times mass\ of\ water$$ $$moles\ of\ urea = 0.100\ \mathrm{m} \times 0.500\ \mathrm{kg}$$ $$moles\ of\ urea = 0.050$$ Answer: There are 0.05 moles of urea in the solution.

Key Concepts

Moles calculation

Calculating the number of moles is an essential skill in chemistry that helps in determining the amount of solute in a solution. Moles are a measurement unit for amount of substance, which is a fundamental concept in chemistry. The formula used to calculate moles is:

• \( ext{moles of solute} = ext{molality} \times ext{mass of solvent (in kg)} \)

Molality (\( ext{m} \)) represents the moles of solute per kilogram of solvent, making it useful in scenarios where temperature changes are involved because it does not depend on volume, which can vary with temperature.

For example, in a \( 0.150 \text{ m} \) glucose solution using \( 100.0 \text{ kg} \) of water, you can find the moles of glucose by multiplying the molality of glucose by the mass of water in kilograms:

• \( 0.150 \text{ m} \times 100.0 \text{ kg} = 15.0 \) moles of glucose

This calculation is straightforward, but is crucial for experiments and reactions where precision is key.

Solution concentration

Solution concentration is a measure of how much solute is present in a given amount of solvent or solution. There are different ways to express concentration, and molality is one of them. It is denoted by the symbol \( ext{m} \) and is defined as the number of moles of solute per kilogram of solvent. Molality is particularly useful when dealing with solutions under changing temperatures because it is based on mass and not volume.

This is different from molarity, which is the number of moles of solute per liter of solution. Molality and molarity are common in stoichiometry, but molality is unique because it ensures consistency across temperature variations. When you are creating or analyzing solutions, knowing how to switch between these concentration units is important for accurate calculations.

• For example, in a \( 0.028 \text{ m} \) \( \text{Na}_2\text{CrO}_4 \) solution, you have \( 0.028 \) moles of \( \text{Na}_2\text{CrO}_4 \) for every kilogram of water.

This consistency in definition helps clarify solution parameters in scientific studies.

Mass conversion

Mass conversion is a basic yet crucial part of solving molality and concentration problems. Often, you will need to convert the mass of a substance from grams to kilograms, as molality calculations rely on the mass of the solvent in kilograms.

• To convert grams to kilograms, simply divide the mass in grams by 1000.

For instance, if you have \( 1000.0 \text{ g} \) of water, convert this to kilograms by doing:

• \( 1000.0 \text{ g} \div 1000 = 1.0 \text{ kg} \)

Using the converted mass helps keep your calculations consistent and accurate, especially in complex reactions.

Let's take another example. For a \( 0.100 \text{ m} \) urea solution with \( 500.0 \text{ g} \) of water, you first convert these grams to kilograms:

• \( 500.0 \text{ g} \div 1000 = 0.500 \text{ kg} \)

Accurate mass conversion allows you to correctly calculate moles and understand the concentration of solutions effectively.