Problem 41 - General Chemistry: Principles and Modern Applications

Question

Which has the higher concentration of sucrose: a \(46 \%\) sucrose solution by mass \((d=1.21 \mathrm{g} / \mathrm{mL}),\) or \(1.50 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) ? Explain your reasoning.

Step-by-Step Solution

Answer

The 1.50 M C12H22O11 solution has a higher sucrose concentration than the 46% sucrose solution by mass.

Step 1: Identify Given Information

In this exercise two different solutions are given. The first solution has 46% sucrose by mass and a density of 1.21 mg/mL. The second solution is 1.50 M sucrose (C12H22O11). The goal is to compare the concentrations of sucrose in these two solutions.

Step 2: Conversion of Percent Solution into Molarity

Firstly, realize that 46% by mass implies that 100 ml solution contains 46 g of sucrose. This is because percentage by mass is defined as the mass of solute per 100 units of solution. So, we first need to calculate how many moles of sucrose (C12H22O11) are present in 46 g. From the molecular formula, the molar mass of sucrose is approximately 342.3 g/mol. Thus, the moles of sucrose can be calculated by dividing the mass by the molar mass: \( \frac{46 g}{342.3 g/mol} = 0.134 mol \). This gives the number of moles in 100 ml, but typically molarity is in moles per liter. Therefore, multiply the result by 10 to convert from 100 ml to 1 L, so the molarity of the first solution is \(0.134 mol/L x 10 = 1.34 M \)

Step 3: Comparison of Molarities

Now compare the molarity of the sucrose in these two solutions. The first solution has a molarity of 1.34 M, while the second solution has a molarity of 1.50 M. Thus, the second solution, 1.50 M C12H22O11, has a higher concentration of sucrose.

Key Concepts

Molarity

Molarity is a common way of expressing the concentration of a solution. It is defined as the number of moles of solute (the substance that is dissolved) per liter of solution. This makes it easy to calculate how much solute is in any given volume of solution. In the context of sucrose, the molecular formula is \( \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), and its molar mass is around 342.3 g/mol.
To find molarity, you typically need to do the following:

• Determine the mass of the solute (e.g., 46 g of sucrose).
• Calculate the moles of the solute using its molar mass: \( \frac{\text{mass of solute}}{\text{molar mass}} \).
• Convert the volume of the solution from milliliters to liters if necessary.
• Divide the moles of solute by the volume in liters to find molarity.

In our exercise, the 1.50 M means there are 1.50 moles of sucrose in every liter of the solution, indicating a comparatively high concentration of sucrose.
Understanding molarity allows for practical applications in fields like chemistry and biology, where precise concentrations are crucial.

Mass Percent

Mass percent is another way of expressing concentration. It tells us how much of a solute is in a solution relative to the total mass of the solution. In other words, it is the ratio of the mass of the solute to the mass of the entire solution, multiplied by 100 to express it as a percentage.
For example, a 46% mass percent sucrose solution means that in every 100 grams of the solution, there are 46 grams of sucrose.
This method is typically used when the density of the solution is known, which allows the concentration to be further converted into other units, such as molarity. This requires the following steps:

• Calculate the mass of the solute based on the mass percent value.
• Use the solution's density to find its total mass.
• Convert the mass of the solute into moles if needed.

In the exercise, the 46% sucrose solution was converted into molarity to compare concentrations effectively. This conversion helps bridge this common mass-based expression to the volume-based expression of molarity, facilitating a straightforward comparison with other solutions.

Solution Density

Solution density is critical when working with mass percent concentrations. Density is defined as the mass of the solution per unit volume, typically expressed in grams per milliliter (g/mL). Knowing the density allows the calculation of mass from a known volume, which is especially useful when you want to convert between mass percent and molarity.
For instance, if the density of the solution is known, as in the case of our exercise with a density of 1.21 g/mL for the sucrose solution, this allows the conversion of a percentage concentration into a molarity. Here's how:

• Using the density to calculate the mass of the solution from its volume.
• Using the mass percent to find out how much of this mass is due to the solute (sucrose).
• Converting this mass to moles of solute using its molar mass.

The density information bridges the gap between knowing how much solute is physically present and understanding how that corresponds to chemical concentrations, like molarity, providing a fuller picture of the solution's makeup.