Problem 51 - Chemistry The Central Science In Si Units

Question

Calculate the number of moles of solute present in each of the following aqueous solutions: (a) \(750 \mathrm{~mL}\) of \(0.120 \mathrm{M}\) \(\operatorname{SrBr}_{2},(\mathbf{b}) 70.0 \mathrm{~g}\) of \(0.200 \mathrm{~m} \mathrm{KCl},(\mathbf{c}) 150.0 \mathrm{~g}\) of a solution that is \(5.75 \%\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) by mass.

Step-by-Step Solution

Answer

(a) 0.090 moles, (b) 0.188 moles, (c) 0.048 moles.

Step 1: Calculate Moles for SrBr2 Solution

For solution (a), use the formula to find moles: \( ext{Moles} = ext{Molarity} \times ext{Volume in Liters} \). Given that the molarity is \( 0.120 \, \text{M} \) and the volume of the solution is \( 750 \, \text{mL} = 0.750 \, \text{L} \), calculate the moles as follows: \( ext{Moles} = 0.120 \, \text{M} \times 0.750 \, \text{L} = 0.090 \, \text{moles} \).

Step 2: Convert Mass to Moles for KCl Solution

For solution (b), first find the molality-related mass of KCl in the solution: \( 70.0 \, \text{g} \times 0.200 \, \text{m} = 14.0 \, \text{g of KCl} \). Convert mass to moles using molar mass of KCl, which is approximately \( 74.55 \, \text{g/mol} \). Moles of KCl = \( \frac{14.0 \, \text{g}}{74.55 \, \text{g/mol}} \approx 0.188 \, \text{moles} \).

Step 3: Calculate Moles for Glucose Solution

For solution (c), calculate the mass of glucose from the percentage by mass: \( 150.0 \, \text{g} \times 0.0575 = 8.625 \, \text{g glucose} \). Convert this mass to moles using the molar mass of glucose, which is \( 180.18 \, \text{g/mol} \). Thus, the moles of glucose are \( \frac{8.625 \, \text{g}}{180.18 \, \text{g/mol}} \approx 0.048 \, \text{moles} \).

Key Concepts

Solution Chemistry

Solution chemistry is all about understanding how substances mix and interact at a molecular level. A solution is formed when a solute, the substance being dissolved, is mixed with a solvent, the substance doing the dissolving. In the examples provided, water serves as the solvent in aqueous solutions because these solutions are water-based. The solutes in our problems include strontium bromide ( SrBr2), potassium chloride ( KCl), and glucose.
Knowing the type and properties of both solute and solvent helps us predict how they will behave together.
Solutions can come in different concentrations, which we commonly express as molarity, molality, or percent composition, depending on what's most useful for our calculations.

Molarity

Molarity ( ect M) is a common way to express the concentration of a solution. It tells us how many moles of solute are present in 1 liter of solution. For example, in our original exercise, we saw that the molarity of the SrBr2 solution was 0.120 M. Here's how molarity allows easy calculations:

• To find moles of a solute: Multiply molarity by the volume of the solution (in liters).
• The formula is: \( ext{Moles} = ext{Molarity} imes ext{Volume in Liters} \)

Using this simple relationship, we can find how many moles are in a solution based on how concentrated it is and its volume. Molarity can vary with temperature because solution volume can expand or contract, so it is important to make sure the conditions are consistent when using and calculating molarity.

Mass to Moles Conversion

Converting mass to moles is often necessary in solution chemistry, especially when you need to understand the amount of a substance present on a molecular scale. The core idea is using the molar mass of a substance, which you can find on the periodic table:

• The molar mass is the weight of 1 mole of a substance, expressed in grams/mole.
• The calculation formula is: \( ext{Moles} = \frac{ ext{Mass in grams}}{ ext{Molar Mass in grams per mole}} \)

In our exercise, this process was used for KCl and glucose. It's crucial to convert from grams to moles because chemical reactions happen on a molecular level, involving amounts measurable in moles.

Concentration Calculations

Concentration calculations are integral to solution chemistry because they provide us with the quantitative relationship between solute and solvent in a solution. With different units such as molarity, molality, and percent composition, we can choose the one that is most applicable:

• Molarity: measures concentration in terms of moles per liter of solution.
• Percent Composition: by mass, shows the percentage of the total mass that is made up of the solute.

For glucose in the exercise, percent composition was used, followed by conversion to moles using the molar mass. This illustrates a multi-step concentration calculation, where multiple conversions are necessary to find the moles of solute. Each concentration unit provides a different view, making concentration calculations versatile tools in chemistry.