Problem 123 - General Chemistry: Principles and Modern Applications

Question

It is desired to produce as large a volume of \(1.25 \mathrm{M}\) urea \(\left[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(\mathrm{aq})\right]\) as possible from these three sources: \(345 \mathrm{mL}\) of \(1.29 \mathrm{M} \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}, 485 \mathrm{mL}\) of \(0.653 \mathrm{M}\) \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2},\) and \(835 \mathrm{mL}\) of \(0.775 \mathrm{M} \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2} .\) How can this be done? What is the maximum volume of this solution obtainable?

Step-by-Step Solution

Answer

The maximum volume of a 1.25 M urea solution that can be obtained from the given sources is 1.13 L.

Step 1: Conversion of milliliters to liters

Since molarity is represented in moles per liter, it's necessary to convert all volumes to liters. 1 L = 1000 mL. Thus, the volumes in liters are: \(345 mL = 0.345 L\), \(485 mL = 0.485 L\), \(835 mL = 0.835 L\).

Step 2: Find the total moles of urea

The number of moles of urea in each solution can be calculated by multiplying the molarity by the volume. Therefore, the total moles of urea is the sum of individual moles from the three sources: \(moles_{total} = 1.29 M * 0.345 L + 0.653 M * 0.485 L + 0.775 M * 0.835 L = 0.44505 moles + 0.316405 moles + 0.646625 moles = 1.40808 moles\).

Step 3: Find the maximum volume of 1.25 M urea solution

Knowing the total number of moles and the desired molarity, the volume can be obtained as \(volume = \frac{moles_{total}}{desired-molarity} = \frac{1.40808 moles}{1.25 M} = 1.126464 L\). The maximum volume of the 1.25 M solution that can be obtained is 1.126464 L, or when rounded to three significant figures, 1.13 L.

Key Concepts

Stoichiometry

Stoichiometry is a branch of chemistry that quantitatively relates the reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of the mole, ensuring that atoms are balanced on both sides of a reaction. Understanding stoichiometry allows for the precise calculation of how much of each reactant is needed to produce a given amount of product, and vice versa.

When applied to our situation, we calculate the maximum volume of 1.25 M urea solution we can derive from different concentrations by using stoichiometry to first determine the total moles of urea available. This is done by multiplying the molarity (the amount of substance per unit volume) by the volume for each solution. We then use the combined moles and the desired molarity to find the final volume of solution.

Solution Concentration

Solution concentration is a measure of how much solute is dissolved in a given amount of solvent. Molarity, one of the ways to express solution concentration, is defined as moles of solute per liter of solution. It's a key concept for preparing solutions in chemistry because it allows chemists to precisely determine how much substance they're working with in a given volume of liquid.

In the exercise, three different urea solutions with varying molarities are combined. To find out how much of a 1.25 M solution can be created from them, we first calculate the number of moles in each given volume and then add them up to get the total moles. The maximum volume of 1.25 M urea solution is found by dividing these total moles by the desired molarity.

Mole Concept

The mole concept is a fundamental principle in chemistry that relates molecular or atomic amounts to a common unit — the mole, which is equivalent to Avogadro's number of entities (approximately 6.022 x 10^23). Understanding the mole concept is essential for translating between microscopic substances, which chemists measure in moles, and macroscopic amounts, which can be measured in grams or liters.

For our exercise, the mole concept helps us understand how we can mix different volumes of urea solutions, with their respective molarity, to obtain a specific concentration (1.25 M). By calculating the total moles of urea from the provided volumes and molarities, we then apply the mole concept again to find out the volume that these moles would occupy if they were in a 1.25 M solution.