Problem 73 - Chemistry The Science in Context

Question

Saccharin Determine the melting point of an aqueous solution made by adding \(186 \mathrm{mg}\) of saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{3} \mathrm{NS}\right)\) to \(\left.1.00 \mathrm{mL} \text { of water (density }=1.00 \mathrm{g} / \mathrm{mL}, K_{f}=1.86^{\circ} \mathrm{C} / \mathrm{m}\right)\)

Step-by-Step Solution

Answer

Answer: The melting point of the aqueous solution is -1.88°C.

Step 1: Calculate the number of moles of saccharin present in the solution

To calculate the number of moles of saccharin, we'll use its molar mass to convert from mass to moles. The molecular formula for saccharin is C7H5O3NS, and the molar mass is C: 12.01 * 7 = 84.07 g/mol H: 1.01 * 5 = 5.05 g/mol O: 16.00 * 3 = 48.00 g/mol N: 14.01 * 1 = 14.01 g/mol S: 32.07 * 1 = 32.07 g/mol Total molar mass = 183.20 g/mol Now, we will use the given mass (186 mg) of saccharin and convert it to moles: moles of saccharin = (186 mg) * (1 g/1000 mg) * (1 mol / 183.20 g) = 0.001015 mol

Step 2: Calculate the mass of water in grams

We will convert the volume of water (1.00 mL) to mass using the given density of water (1.00 g/mL). mass of water = (1.00 mL) * (1.00 g/mL) = 1.00 g

Step 3: Calculate the molality of the solution

Molality is defined as moles of solute per kilogram of solvent. We will use the moles of saccharin and mass of water to find the molality of the solution: molality (m) = (0.001015 mol) / (1.00 g * 1 kg/1000 g) = 1.015 mol/kg

Step 4: Calculate the change in the freezing point

We will use the freezing point depression formula (ΔTf = Kf * m) to calculate the change in freezing point. The given cryoscopic constant (Kf) of water is 1.86°C/m. ΔTf = (1.86°C/m) * (1.015 mol/kg) = 1.88°C

Step 5: Determine the new melting point of the aqueous solution

To find the new melting point of the solution, we will subtract the change in freezing point from the normal freezing point of water (0°C). New melting point = 0°C - 1.88°C = -1.88°C So, the melting point of an aqueous solution made by adding 186 mg of saccharin to 1.00 mL of water is -1.88°C.

Key Concepts

Molality

Molality is a way to express the concentration of a solution. It's especially useful when dealing with changes in temperature. Unlike molarity, which depends on the volume of the solution, molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent:\[\text{molality} (m) = \frac{\text{moles of solute}}{\text{kg of solvent}}\]In the exercise, the molality of the solution was calculated by using the moles of saccharin and the mass of water. This value is crucial for determining the change in freezing point of the solution. The advantage of using molality is that it doesn't change with temperature, making it reliable for problems involving temperature changes.

Molar Mass

Understanding molar mass is key when converting between the mass of a substance and the amount in moles.The molar mass of a compound is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance.For saccharin, the molar mass was calculated by summing the atomic masses of all the atoms in its molecular formula (\(\text{C}_{7}\text{H}_{5}\text{O}_{3}\text{NS}\)):

• Carbon (C): 12.01 g/mol × 7 = 84.07 g/mol
• Hydrogen (H): 1.01 g/mol × 5 = 5.05 g/mol
• Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol
• Nitrogen (N): 14.01 g/mol × 1 = 14.01 g/mol
• Sulfur (S): 32.07 g/mol × 1 = 32.07 g/mol

This results in a total molar mass of 183.20 g/mol for saccharin. This value was then used to convert the mass of saccharin into moles. Having the molar mass allows chemists to scale their reactions accurately.

Aqueous Solution

An aqueous solution is simply a solution in which the solvent is water. Many reactions and processes in chemistry occur in aqueous solutions because water is a universal solvent. In this exercise, saccharin was dissolved into water to form an aqueous solution. This involved measuring an exact amount of water, in this case, 1.00 mL, which was then converted into grams by using the known density of water (1.00 g/mL). Aquatic solutions are essential when studying properties like freezing point depression as they directly influence the final result.

Cryoscopic Constant

The cryoscopic constant (\(K_f\)) is a property of the solvent that indicates how much the freezing point is lowered when a solute is added. The constant is specific to each solvent. For water, the cryoscopic constant is given as 1.86°C/m.Utilizing the cryoscopic constant in the freezing point depression formula allows us to calculate how much the freezing point is lowered:\[\Delta T_f = K_f \times m\]Where:\(\Delta T_f\) is the change in freezing point,\(K_f\) is the cryoscopic constant, and\(m\) is the molality of the solution.This relationship is essential in predicting the behavior of solutions under different thermal conditions. By knowing how a solute affects the freezing point, we can better understand solution dynamics.