Question

A solution is prepared by dissolving 28.4 \(\mathrm{g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in 355 g of water. The final volume of the solution is 378 \(\mathrm{mL}\) . For this solution, calculate the concentration in each unit. a. molarity b. molality c. percent by mass d. mole fraction e. mole percent

Short answer

Molarity = \( \frac{0.158 \mathrm{moles}}{0.378 \mathrm{L}} \) = 0.417 M, Molality = \( \frac{0.158 \mathrm{moles}}{0.355 \mathrm{kg}} \) = 0.445 m, Percent by Mass = \( \frac{28.4 \mathrm{g}}{383.4 \mathrm{g}} \times 100\text{%} \) = 7.41%, Mole Fraction of Glucose ≈ 0.0099, Mole Percent of Glucose ≈ 0.99%.

Step-by-step solution

Calculate Moles of Glucose

To calculate molarity, molality, and other concentration terms, begin by calculating the moles of glucose. Use its molar mass to convert grams to moles. The molar mass of glucose, \( \mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6} \) is \( 6 \times 12.01 + 12 \times 1.008 + 6 \times 16.00 = 180.16 \mathrm{g/mol} \). The number of moles of glucose is therefore \[ \frac{28.4 \mathrm{g}}{180.16 \mathrm{g/mol}} \].

Calculate Molarity (M)

Molarity (M) is moles of solute per liter of solution. Convert the volume from milliliters to liters by dividing by 1000, and then use the moles of glucose to find molarity. \[ \text{Molarity} = \frac{\text{Moles of Glucose}}{\text{Volume of Solution in Liters}} \]

Calculate Molality (m)

Molality (m) is moles of solute per kilogram of solvent. Convert the mass of water from grams to kilograms by dividing by 1000, and then use the moles of glucose to find molality. \[ \text{Molality} = \frac{\text{Moles of Glucose}}{\text{Mass of Water in Kilograms}} \]

Calculate Percent by Mass (%)

Percent by mass is the mass of the solute divided by the total mass of the solution, multiplied by 100. Add the mass of glucose to the mass of water to get the total mass, then find the percent by mass. \[ \text{Percent by Mass} = \left( \frac{\text{Mass of Glucose}}{\text{Total Mass of Solution}} \right) \times 100\text{%} \]

Calculate Mole Fraction of Glucose

The mole fraction is the moles of solute divided by the total moles of all components. You'll also need to calculate moles of water using its molar mass (18.01528 g/mol). After that, use both moles of glucose and water to find the mole fraction of glucose. \[ \text{Mole Fraction of Glucose} = \frac{\text{Moles of Glucose}}{\text{Moles of Glucose} + \text{Moles of Water}} \]

Calculate Mole Percent of Glucose

Mole percent is the mole fraction multiplied by 100. Once you have the mole fraction of glucose from Step 5, multiply it by 100 to get the mole percent. \[ \text{Mole Percent of Glucose} = \text{Mole Fraction of Glucose} \times 100\text{%} \]

Key concepts

Molarity

Understanding molarity is essential when discussing solutions. Molarity, often represented by the symbol 'M', quantifies the number of moles of a solute present in one liter of solution. In simpler terms, it tells you the concentration or strength of a solution.

For example, if you dissolve a certain amount of substance in water, the molarity tells you how many moles of this substance you'll find in each liter. This is especially useful in chemistry experiments where reaction rates are determined by the concentration of the reactants.

When calculating molarity, remember it's about the total volume of the solution, not just the volume of the solvent. This is a common point where mistakes are made, so ensure the volume you use includes the solute and the solvent.

Molality

Molality represents another way to express the concentration of a solution. While molarity talks about moles per liter of solution, molality focuses on moles per kilogram of solvent. Denoted by 'm', this unit isn't affected by changes in temperature or pressure, unlike molarity. This makes molality particularly valuable for studies involving varying environmental conditions.

The calculation for molality leans heavily on the mass of the solvent, requiring you to think in kilograms rather than liters. One important scenario where molality is favored is when dealing with boiling point elevation and freezing point depression, which are colligative properties that rely on the number of particles in a solvent.

Percent by Mass

The percent by mass, sometimes indicated as weight percent or mass percent, offers a simple perspective on solution concentration. It represents the mass of the solute as a percentage of the total mass of the solution. This measure is intuitive and particularly useful when the precise volume of the solution isn't easy to determine.

Calculating percent by mass is straightforward: you only need to know the mass of your solute and the total mass of the solution, which includes the mass of the solute plus the mass of the solvent. It's a useful metric when preparing solutions in industries like pharmaceuticals or cooking where precision based on weight is critical.

Mole Fraction

The mole fraction is a way to express the ratio of the number of moles of one component to the total number of moles in a mixture. Because it is a ratio, it has no units and is useful for comparing the amount of one substance to the whole without worrying about the volume or mass of the solution.

Calculating the mole fraction involves knowledge of all components in the solution, not just a single solute. This measure is quite handy in gas mixtures and when dealing with Raoult's Law which correlates to vapor pressure of solutions.

Mole Percent

Closely related to the mole fraction is the mole percent. The mole percent takes the mole fraction and multiplies it by 100 to express the concentration as a percentage. In essence, it tells you how many moles out of 100 moles are of the interest substance.

This unit of measurement can be particularly useful when you're dealing with multiple solutes in the same solvent, as it can give you an instant idea of the 'popularity' of each solute in the crowd of the solution. It's like stating the percentage of a certain type of people in a large group, making it clear and easy to understand.