Problem 45 - Chemistry and Chemical Reactivity

Question

The melting point of pure biphenyl \(\left(\mathrm{C}_{12} \mathrm{H}_{10}\right)\) is found to be \(70.03^{\circ} \mathrm{C} .\) If \(0.100 \mathrm{g}\) of naphthalene is added to \(10.0 \mathrm{g}\) of biphenyl, the freezing point of the mixture is \(69.40^{\circ} \mathrm{C} .\) If \(K_{\mathrm{fp}}\) for biphenyl is \(-8.00^{\circ} \mathrm{C} / \mathrm{m},\) what is the molar mass of naphthalene?

Step-by-Step Solution

Answer

The molar mass of naphthalene is 127 g/mol.

Step 1: Identify the formulas involved

We will first identify the formulas needed to solve the problem. The freezing point depression formula is given by:\[\Delta T_{fp} = i \cdot K_{fp} \cdot m\]where \(\Delta T_{fp}\) is the freezing point depression, \(i\) is the van't Hoff factor (which is 1 for non-electrolytes such as naphthalene), \(K_{fp}\) is the freezing point depression constant, and \(m\) is the molality of the solution.

Step 2: Calculate the freezing point depression

The depression in freezing point \(\Delta T_{fp}\) can be found by subtracting the freezing point of the mixture from that of pure biphenyl:\[\Delta T_{fp} = 70.03^{\circ}C - 69.40^{\circ}C = 0.63^{\circ}C\]

Step 3: Calculate the molality

The molality \(m\) of the solution can be found by rearranging the freezing point depression formula:\[m = \frac{\Delta T_{fp}}{i \cdot K_{fp}} = \frac{0.63^{\circ}C}{1 \cdot -8.00^{\circ}C/m} = 0.07875 \,m\]

Step 4: Calculate the moles of solute

Molality is defined as moles of solute per kilogram of solvent. We rearrange the definition to find the moles of naphthalene:\[\text{Moles of naphthalene} = m \times \text{mass of biphenyl in kg}\]\[= 0.07875 \,m \times 0.010 \text{ kg} = 7.875 \times 10^{-4} \text{ moles}\]

Step 5: Calculate the molar mass of naphthalene

Finally, the molar mass \( M \) of naphthalene can be calculated by dividing the mass of naphthalene by the moles we just found:\[M = \frac{0.100 \text{ g}}{7.875 \times 10^{-4} \text{ moles}} = 127 \text{ g/mol}\]

Key Concepts

Molar Mass Calculation

Calculating the molar mass is like trying to find the weight of a single characteristic unit of a substance, known as the molecule. This involves understanding how much an individual mole, or a standard number of particles, weighs in grams. To determine the molar mass of naphthalene from our example, we start with a measured mass of the naphthalene sample.
The key formula used is:

• \( M = \frac{\text{mass of naphthalene}}{\text{moles of naphthalene}} \)

To find the number of moles, we use the relationship derived from our knowledge of freezing point depression, which tells us how the addition of a solute (like naphthalene) affects the freezing point of the solvent (biphenyl).
This calculated mass per mole expresses the molar mass as grams per mole (g/mol), a useful measure for understanding molecular scale.

Van't Hoff Factor

The van’t Hoff factor, symbolized by \( i \), is a crucial concept in understanding how solutes impact the properties of solutions. Essentially, it describes the effect of solute particles on the colligative properties of a solution, like freezing point depression.

In the context of our exercise, naphthalene is classified as a nonelectrolyte. This means it does not dissociate into ions when dissolved. Thus, the van't Hoff factor \( i \) becomes 1 because each naphthalene molecule contributes exactly one particle to the solution.

• For nonelectrolytes, \( i = 1 \).
• For electrolytes, \( i \) would represent the number of particles the solute dissociates into. For example, \( NaCl \) with \( i = 2 \).

In summary, understanding the van’t Hoff factor helps predict how solutes alter physical properties, crucial for processes like calculating freezing point depression.

Molality

Molality is a way to express the concentration of a solution, which is different from other concentration measures like molarity. It's crucial for calculations involving temperature-dependent phenomena.
Molality is defined as the number of moles of solute per kilogram of solvent, not the total solution. Its usefulness shines in colligative property calculations, like determining how much a solute will depress the freezing point of a solvent.
To calculate molality, we use:

• \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)

In our problem, after finding the change in freezing point, we rearrange the freezing point depression formula to solve for molality. This strategic approach helps us understand solutions on a microscopic level, revealing the strength of solute-solvent interactions.