Problem 85 - Chemistry: Principles and Reactions

Question

The water-soluble nonelectrolyte \(\mathrm{X}\) has a molar mass of \(410 \mathrm{~g} / \mathrm{mol}\). A \(0.100\) -g mixture containing this substance and sugar \((\mathrm{MM}=342 \mathrm{~g} / \mathrm{mol})\) is added to \(1.00 \mathrm{~g}\) of water to give a solution whose freezing point is \(-0.500^{\circ} \mathrm{C}\). Estimate the mass percent of \(\mathrm{X}\) in the mixture.

Step-by-Step Solution

Answer

Question: Estimate the mass percent of substance X in a mixture containing substance X and sugar given that the molar mass of substance X is 410 g/mol, the molar mass of sugar is 342 g/mol, the mass of the mixture is 0.100 g, and the resulting solution's freezing point is -0.500°C. Answer: The mass percent of substance X in the mixture is approximately 43.2%.

Step 1: Calculate the freezing-point depression constant (Kf) for water

Knowing that the freezing point of pure water is \(0^{\circ} \mathrm{C}\), and the freezing point of the solution is \(-0.500^{\circ} C\), we can calculate the freezing-point depression \(\Delta T_f\) as: \(\Delta T_f = T_f^\text{pure} - T_f^\text{solution} = 0 - (-0.500) = 0.500^{\circ} \mathrm{C}\) For water, the freezing-point depression constant (\(K_f\)) is \(1.86 \, \mathrm{K \, kg/mol}\). Now, we can find the molality (m) of solute using the formula: \(\Delta T_f = K_f \cdot m\) Rearranging to solve for molality (m): \(m = \frac{\Delta T_f}{K_f} = \frac{0.500^{\circ} \mathrm{C}}{1.86 \, \mathrm{K \, kg/mol}} = 0.2691 \, \mathrm{mol/kg}\)

Step 2: Calculate the moles of solute in the solution

Since total mass of the mixture (\(0.100 \, \mathrm{g}\)) was added to \(1.00 \, \mathrm{g}\) of water, the total mass of the solution is \(1.10 \, \mathrm{g}\). Molality (m) is the number of moles of solute per kilogram solute. We can use molality, total mass of the solution, and given molar masses to calculate the moles of substance X and sugar in the solution. Let \(w_X\) be the mass of substance X in the mixture. Then, the mass of sugar in the mixture is \((0.100 - w_X)\, \mathrm{g}\). Substance X: \(n_X = \frac{w_X}{\mathrm{MM_X}}\) Sugar: \(n_S = \frac{0.100 - w_X}{\mathrm{MM_S}}\) Since the molality of the solution is the sum of the molalities of substance X and sugar: \(m = 0.2691 \, \mathrm{mol/kg} = \frac{n_X}{w_w} + \frac{n_S}{w_w}\) Replacing moles and masses in terms of \(w_X\): \(0.2691 \, \mathrm{mol/kg} = \frac{\frac{w_X}{\mathrm{MM_X}}}{w_w} + \frac{\frac{0.100 - w_X}{\mathrm{MM_S}}}{w_w}\)

Step 3: Solve for the mass of substance X (wX) and calculate the mass percent

We can solve for \(w_X\) to find the mass of substance X in the mixture: \(0.2691 \, \mathrm{mol/kg} = w_X\,(\frac{1}{410\, \mathrm{g/mol}}-\frac{1}{342\, \mathrm{g/mol}})\frac{1}{1.00\, \mathrm{g}}\) Now, isolate \(w_X\): \(w_X = \frac{0.2691 \, \mathrm{mol/kg}}{(\frac{1}{410\, \mathrm{g/mol}}-\frac{1}{342\, \mathrm{g/mol}})\frac{1}{1.00\, \mathrm{g}}} \approx 0.0432\, \mathrm{g}\) Finally, we can calculate the mass percent of substance X in the mixture: Mass Percent \(= \frac{0.0432\, \mathrm{g}}{0.100\, \mathrm{g}} \times 100\% \approx 43.2\%\) Thus, the mass percent of substance X in the mixture is approximately 43.2%.

Key Concepts

Molality

Molality is a measure of the concentration of a solute in a solution. It is expressed as moles of solute per kilogram of solvent, rather than per liter of solution like molarity. This makes molality useful for applications relating to temperature changes, such as freezing point depression and boiling point elevation. Unlike molarity, molality is not affected by temperature changes since it is based on mass, not volume.

When calculated, molality can help us understand how a solute affects the properties of a solvent. It’s calculated using the formula: \\[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \]

In the exercise, the freezing point depression was used to calculate the molality of the solution containing both the nonelectrolytic substance X and sugar. The formula used was \\( \Delta T_f = K_f \cdot m \), where \\( \Delta T_f \) is the change in freezing point, and \\( K_f \) is the freezing-point depression constant.

Molar Mass

Molar mass is the mass of one mole of a given substance and is expressed in grams per mole (g/mol). It’s a fundamental property of a chemical compound, connecting the number of moles to the mass and crucial in calculations involving chemical reactions.

For instance, in the given problem, the molar masses of substances X and sugar were given as 410 g/mol and 342 g/mol, respectively. These values allow us to determine how many moles of each substance are present in the mixture by using the formula:
\[ n = \frac{m}{M} \]
where \\( n \) is the number of moles, \\( m \) is the mass, and \\( M \) is the molar mass.

By knowing molar masses, one can calculate the amount of substance present and hence further calculate the resulting concentration or the effect on colligative properties like freezing point depression.

Mass Percent Calculation

Mass percent, also called mass percent composition, is a way of expressing the concentration of an element in a compound or a component in a mixture. It is calculated as the mass of the component divided by the total mass of the mixture, multiplied by 100 to express it as a percentage.

In our example of estimating the mass percent of substance X in a sugar mixture, the formula used is:

• Mass percent = \( \frac{\text{mass of component}}{\text{total mass of mixture}} \times 100 \% \)

By substituting the specific masses, students can find the mass percent of each component. The calculated mass percent informs us of how much of the total mixture is made up by each component. This can help in understanding the contribution of each solute to the overall properties of the solution.

Nonelectrolyte

A nonelectrolyte is a substance that, when dissolved in water, does not dissociate into ions. This means that nonelectrolytes do not conduct electricity in their aqueous solutions. Examples of nonelectrolytes include sugar and many organic compounds like alcohol.

In the context of freezing point depression, nonelectrolytes cause changes in the physical properties of solutions but do not affect the electrical conductivity. These substances exert influence on colligative properties, such as osmotic pressure, boiling point elevation, and freezing point depression.

The nonelectrolytic property of substances like X is significant when calculating colligative properties. Since these properties depend on the number of solute particles, and not their identity, knowing whether a substance is a nonelectrolyte helps in predicting its effect on the solution's freezing point.