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A solution contains \(3.75 \mathrm{g}\) of a nonvolatile pure hydrocarbon in \(95 \mathrm{g}\) acetone. The boiling points of pure acetone and the solution are \(55.95^{\circ} \mathrm{C}\) and \(56.50^{\circ} \mathrm{C},\) respectively. The molal boilingpoint constant of acetone is \(1.71^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} .\) What is the molar mass of the hydrocarbon?

Short Answer

Expert verified
The molar mass of the hydrocarbon is approximately 122.7 g/mol.

Step by step solution

01

Calculate the boiling point elevation

Boiling point elevation (ΔT_b) is the difference between the boiling point of the solution and the boiling point of the pure solvent. In this case, the solvent is acetone. ΔT_b = T_solution - T_solvent = 56.50°C - 55.95°C = 0.55°C
02

Calculate the molality of the hydrocarbon

We can use the formula ΔT_b = Kb * molality to find the molality of the hydrocarbon in the solution. Molality = ΔT_b / Kb = 0.55°C / 1.71°C·kg/mol = 0.3216 mol/kg
03

Calculate the moles of hydrocarbon in the solution

Since molality represents the moles of solute per kilogram of solvent, we can calculate the moles of hydrocarbon, knowing the mass of acetone in the solution: moles_hydrocarbon = molality * mass_of_acetone / 1000 = 0.3216 mol/kg * 95g * (1 kg/1000g) = 0.03054 mol
04

Calculate the mass of hydrocarbon per mole

We are given the mass of the hydrocarbon in the solution, so we can calculate the mass of hydrocarbon per mole: mass_per_mole = mass_of_hydrocarbon / moles_hydrocarbon = 3.75g / 0.03054 mol = 122.7 g/mol
05

Determine the molar mass of the hydrocarbon

The molar mass of the hydrocarbon is the mass per mole calculated in the previous step. Molar mass of hydrocarbon = 122.7 g/mol The molar mass of the hydrocarbon is approximately 122.7 g/mol.

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Key Concepts

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

Molar Mass
Molar mass is a fundamental concept in chemistry that refers to the mass of a given substance (element or compound) divided by the amount of substance. It is expressed in grams per mole (g/mol), representing the weight of one mole of molecules or atoms of a material. Calculating molar mass is essential for understanding how substances interact in chemical reactions, as it determines the proportions in which different materials combine.

In the context of the problem, finding the molar mass involves determining the mass of the hydrocarbon present and dividing it by the number of moles. Since we have the mass of the solute as 3.75 g and moles calculated in step 4 as 0.03054 mol, we divide these values to find the molar mass to be about 122.7 g/mol. Understanding molar mass is crucial for comprehending the physical and chemical properties of substances.
  • Helps quantify chemical reactions
  • Assists in converting moles to grams
  • Essential for calculating concentrations
Molality
Molality ( μ ) is another way to express the concentration of a solution, defined as the number of moles of solute per kilogram of solvent. Unlike molarity, molality is not dependent on temperature since it relies solely on the mass for its calculation rather than volume, which can change with temperature.

In the given problem, molality is calculated using the boiling point elevation formula, ΔT_b = K_b × molality, where the molal boiling point constant ( K_b ) is a unique value for each solvent. For acetone, K_b is 1.71°C·kg/mol. Given the boiling point elevation, ΔT_b, as 0.55°C, we find that the molality of the hydrocarbon is 0.3216 mol/kg. This concept illustrates how changes in boiling point can help determine concentration values of solutes in a solution.
  • Independent of temperature
  • Utilizes mass to measure concentration
  • Useful for boiling and freezing point calculations
Hydrocarbon
Hydrocarbons are organic compounds composed entirely of hydrogen and carbon atoms. They are the simplest type of organic compound and serve as the backbone for more complex molecules in organic chemistry. Hydrocarbons are classified based on the types of bonds between carbon atoms:

1. Alkanes (single bonds)
2. Alkenes (double bonds)
3. Alkynes (triple bonds)

In this exercise, the hydrocarbon is a nonvolatile solute in acetone, affecting the solution's boiling point, which allows us to calculate properties like molality and molar mass. Understanding hydrocarbons is vital because they:
  • Form the basis for all organic chemistry
  • Are major components of fuels and plastics
  • Vary significantly in their physical and chemical properties based on structure
Acetone
Acetone is a simple ketone with the molecular formula C_3H_6O. It is a clear, volatile, flammable liquid commonly used as a solvent in cleaning and nail polish removers. Its properties as a solvent make it particularly important in both industrial and laboratory settings.

Acetone's ability to dissolve a wide range of substances, and its high volatility, mean that when a solute like a hydrocarbon is added, significant changes can occur in properties like the boiling point. In this particular situation, acetone aids in determining the molar mass and molality of a solute due to its known boiling point elevation constant. Acetone's significance is highlighted by its role in scientific calculations involving solution properties, emphasizing its practical applications in chemistry research and industry.
  • Highly effective solvent
  • Commonly used in industry
  • Useful in calculating changes in solution properties

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Most popular questions from this chapter

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(25^{\circ} \mathrm{C}\) ) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C}\). You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C}\) ) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at \(\left.25^{\circ} \mathrm{C}\right)\) that has a boiling point of \(102.0^{\circ} \mathrm{C}\). What will happen to the plant cells in the leaf?

You have read that adding a solute to a solvent can both increase the boiling point and decrease the freezing point. A friend of yours explains it to you like this: "The solute and solvent can be like salt in water. The salt gets in the way of freezing in that it blocks the water molecules from joining together. The salt acts like a strong bond holding the water molecules together so that it is harder to boil." What do you say to your friend?

Patients undergoing an upper gastrointestinal tract laboratory test are typically given an X-ray contrast agent that aids with the radiologic imaging of the anatomy. One such contrast agent is sodium diatrizoate, a nonvolatile water-soluble compound. A \(0.378-m\) solution is prepared by dissolving 38.4 g sodium diatrizoate (NaDTZ) in \(1.60 \times 10^{2} \mathrm{mL}\) water at \(31.2^{\circ} \mathrm{C}\) (the density of water at \(31.2^{\circ} \mathrm{C}\) is \(0.995 \mathrm{g} / \mathrm{cm}^{3}\) ). What is the molar mass of sodium diatrizoate? What is the vapor pressure of this solution if the vapor pressure of pure water at \(31.2^{\circ} \mathrm{C}\) is 34.1 torr?

A solution of phosphoric acid was made by dissolving \(10.0 \mathrm{g}\) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in \(100.0 \mathrm{mL}\) water. The resulting volume was \(104 \mathrm{mL}\) Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of \(1.00 \mathrm{g} / \mathrm{cm}^{3}\).

Which of the following will have the lowest total vapor pressure at \(25^{\circ} \mathrm{C} ?\) a. pure water (vapor pressure \(=23.8\) torr at \(25^{\circ} \mathrm{C}\) ) b. a solution of glucose in water with \(\chi_{\mathrm{C}_{\mathrm{s}} \mathrm{H}_{\mathrm{l} 2} \mathrm{O}_{\mathrm{s}}}=0.01\) c. a solution of sodium chloride in water with \(\chi_{\mathrm{NaCl}}=0.01\) d. a solution of methanol in water with \(\chi_{\mathrm{CH_{3}}, \mathrm{OH}}=0.2\) (Consider the vapor pressure of both methanol \([143\) torr at \(\left.25^{\circ} \mathrm{C}\right]\) and water.

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