Solutions Lecture 3 Notes

Mole Fraction

Definition: Mole fraction is a dimensionless quantity representing the ratio of the number of moles of a specific component to the total number of moles of all components in a solution. This ratio is crucial in understanding the composition of mixtures and plays a significant role in various fields, such as chemistry and engineering. The mole fraction ranges from 0 to 1, where 0 indicates that the component is absent, and 1 indicates it is the only component present.

Formula:
\text{Mole Fraction of Component} = \frac{\text{Moles of Component}}{\text{Total Moles of Solution}}

Example - Iodine and Methylene Chloride

Given:

• Iodine: 1.50 grams

• Methylene Chloride (CH₂Cl₂): 56.0 grams

1. Calculate Moles of Iodine:
   Molar mass of iodine (I): approximately 126.9 g/mol
   \text{Moles of Iodine} = \frac{1.50 \text{ g}}{126.9 \text{ g/mol}} = 0.0118 \text{ moles} (Rounded to 4 significant figures)

2. Calculate Moles of Methylene Chloride:
   Molar mass of methylene chloride: approximately 84.93 g/mol
   \text{Moles of Methylene Chloride} = \frac{56.0 \text{ g}}{84.93 \text{ g/mol}} = 0.6588 \text{ moles}

3. Calculate Mole Fraction of Iodine:
   Total moles = moles of iodine + moles of methylene chloride
   \text{Total Moles} = 0.0118 + 0.6588 = 0.6706 \text{ moles}
   \text{Mole Fraction of Iodine} = \frac{0.0118}{0.6706} = 0.0176

4. Calculate Mole Fraction of Methylene Chloride:
   Utilizing the previously calculated mole fraction of iodine,
   \text{Mole Fraction of Methylene Chloride} = 1 - \text{Mole Fraction of Iodine} = 1 - 0.0176 = 0.9824

Converting Concentration Units

Converting Percent by Mass to Molarity

Given a 10% glucose (C₆H₁₂O₆) solution: 10.0 g of glucose in 100 g of solution.

Goal: Convert percent by mass to molarity, a vital process in solution preparation in laboratory settings.

1. Determine Moles of Glucose:
   Molar mass of glucose: approximately 180 g/mol
   \text{Moles of Glucose}=\frac{10.0 \text{ g}}{180 \text{ g/mol}}=0.0556\text{ moles} (Rounded to 4 significant figures)

2. Convert Solution Mass to Volume:
   To find the volume of the solution, we use the density: 1.04 g/mL.
   \text{Volume of solution} = \frac{100 \text{ g}}{1.04 \text{ g/mL}} = 96.15 \text{ mL} = 0.09615 \text{ L}

3. Calculate Molarity (M):
   The relationship of molarity is given by the formula: M = \frac{\text{Moles of Solute}}{\text{Volume of Solution in Liters}}
   M=\frac{0.0556\text{ moles}}{0.09615\text{ L}}=0.578\text{ M}
   This indicates that there are 0.578 moles of glucose in each liter of solution.

Converting Percent by Mass to Molality

1. Start with Composition:
   In a typical 10% glucose solution, this means there are 10.0 g of glucose in 100 g of solution, of which 90 g is the solvent (water).

2. Convert Grams to Kilograms:
   90 \text{ g} = 0.090 \text{ kg}
   It is essential to use kilograms for calculating molality.

3. Calculate Moles of Glucose (Already done):
   0.0556\text{ moles}

4. Calculate Molality (m):
   The formula for molality is given by: m = \frac{\text{Moles of Solute}}{\text{Kilograms of Solvent}}
   m=\frac{0.0556\text{ moles}}{0.090\text{ kg}}=0.618\text{ m}
   This value indicates the concentration of glucose in terms of moles of solute per kilogram of solvent.

Conclusion

In summary, mole fractions are crucial in describing the proportions of components within solutions, providing insight into their behavior and reactions. Additionally, converting between concentration units such as molarity and molality requires a solid understanding of the relationships between the components of a solution, their amounts, and their physical properties (mass, volume, density). Applying step-by-step procedures and practicing these calculations are essential for mastering the conversion process and effectively preparing solutions in various scientific contexts!