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Ascorbic acid (vitamin \(\mathrm{C}_{,} \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\) ) is a water-soluble vitamin. A solution containing \(80.5 \mathrm{~g}\) of ascorbic acid dissolved in \(210 \mathrm{~g}\) of water has a density of \(1.22 \mathrm{~g} / \mathrm{mL}\) at \(55^{\circ} \mathrm{C}\). Calculate (a) the mass percentage, (b) the mole fraction, (c) the molality, (d) the molarity of ascorbic acid in this solution.

Short Answer

Expert verified
The following are the properties of ascorbic acid (C6H8O6) in the given solution: (a) Mass percentage ≈ 27.72% (b) Mole fraction of ascorbic acid ≈ 0.0377, and water ≈ 0.9623 (c) Molality ≈ 2.18 mol/kg (d) Molarity ≈ 1.92 mol/L

Step by step solution

01

Find the mass of ascorbic acid and water in the solution

Given information: - Mass of ascorbic acid (m_ascorbic) = 80.5 g - Mass of water (m_water) = 210 g
02

Calculate the Molar mass of ascorbic acid

Using the chemical formula (C6H8O6), we calculate the molar mass of ascorbic acid as follows: Molar mass = (6 * 12.01) + (8 * 1.01) + (6 * 16.00) = 72.06 + 8.08 + 96.00 = 176.14 g/mol
03

Calculate the Mass percentage of ascorbic acid in the solution

We can now calculate the mass percentage of ascorbic acid in the solution using the formula: Mass percentage = (mass of ascorbic acid / total mass) * 100 = (80.5 g / (80.5 g + 210 g)) * 100 ≈ 27.72%
04

Calculate the mole fraction of ascorbic acid and water in the solution

First, we need to find the moles of ascorbic acid and water: Moles of ascorbic acid (n_ascorbic) = mass / molar mass = 80.5 g / 176.14 g/mol ≈ 0.457 moles For water, the molar mass is approximately 18.01 g/mol: Moles of water (n_water) = mass / molar mass = 210 g / 18.01 g/mol ≈ 11.66 moles Now we can compute the mole fraction for both: Mole fraction of ascorbic acid (X_ascorbic) = n_ascorbic / (n_ascorbic + n_water) ≈ 0.457 / (0.457 + 11.66) ≈ 0.0377 Mole fraction of water (X_water) = n_water / (n_ascorbic + n_water) ≈ 11.66 / (0.457 + 11.66) ≈ 0.9623
05

Calculate the molality of the ascorbic acid solution

Now we can find the molality of the solution as follows: Molality = moles of ascorbic acid / kg of water ≈ 0.457 moles / 0.210 kg ≈ 2.18 mol/kg
06

Calculate the molarity of the ascorbic acid solution

First, we need to find the volume of the solution using the given density: density = mass / volume volume = mass / density = (80.5 g + 210 g) / 1.22 g/mL ≈ 238.03 mL Now convert the volume to liters and compute the molarity: Volume (in L) = 238.03 mL / 1000 ≈ 0.238 L Molarity = moles of ascorbic acid / volume in liters ≈ 0.457 moles / 0.238 L ≈ 1.92 mol/L #Summary of Results# a. Mass percentage of ascorbic acid ≈ 27.72% b. Mole fraction: - Ascorbic acid: ≈ 0.0377 - Water: ≈ 0.9623 c. Molality of ascorbic acid ≈ 2.18 mol/kg d. Molarity of ascorbic acid ≈ 1.92 mol/L

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Percentage
Understanding the mass percentage of a solute in a solution is crucial as it directly correlates with the purity and concentration of the substance. Mass percentage is a measure of concentration that expresses the amount of solute (in this case, ascorbic acid) as a percentage of the total mass of the solution. The formula to calculate mass percentage is:
\[ \text{Mass percentage} = \left( \frac{\text{mass of the solute}}{\text{total mass of the solution}} \right) \times 100 \% \] This straightforward formula requires you to sum the mass of the solute and solvent to determine the total mass before working the calculation. With these basics, students can apply this concept to determine the purity level of samples in various scientific and industrial settings.
Mole Fraction
The mole fraction is another way to describe the concentration of a component in a mixture. It is dimensionless, which means it has no units, as it represents the ratio of the number of moles of one component (ascorbic acid in our exercise) to the total number of moles of all components in the mixture. Specifically, the mole fraction is useful when we are dealing with gas mixtures or when expressing the composition of liquid solutions in chemistry. The way to calculate the mole fraction is simply:
\[ \text{Mole fraction} = \frac{\text{moles of the component}}{\text{total moles in the mixture}} \] By applying this concept, one can better understand the mixing ratios and molar relationships in chemical reactions, making it a fundamental concept in chemical stoichiometry.
Molality
Molality measures concentration, similar to molarity, but with a key difference. It is defined as the moles of solute per kilogram of solvent, which makes it particularly useful in scenarios where temperature changes are involved. Since mass does not change with temperature, unlike volume, molality is unaffected by temperature fluctuations. The calculation for molality can be expressed as:
\[ \text{Molality} = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \] It's beneficial when studying properties that depend on the concentration of solute particles but are independent of temperature, such as boiling point elevation or freezing point depression. Students will find a firm grasp of molality useful in thermodynamics and physical chemistry.
Molarity
Molarity is one of the most common units of concentration used in chemistry. It is defined as the number of moles of a solute dissolved per liter of solution. Unlike molality, molarity is temperature-dependent since it involves volume, which can expand or contract with temperature. The formula to calculate molarity is given by:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \] Grasping the concept of molarity is important for students to proceed with quantitative aspects of chemical reactions, such as determining reactant or product quantities based on reaction stoichiometry. Additionally, in laboratory settings, molarity is often used to prepare solutions with precise solute concentrations.

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Most popular questions from this chapter

A saturated solution of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is made by dissolving excess table sugar in a flask of water. There are \(50 \mathrm{~g}\) of undissolved sucrose crystals at the bottom of the flask in contact with the saturated solution. The flask is stoppered and set aside. A year later a single large crystal of mass \(50 \mathrm{~g}\) is at the bottom of the flask. Explain how this experiment provides evidence for a dynamic equilibrium between the saturated solution and the undissolved solute.

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