Caryophyllene: A Nonelectrolyte Is One Of The Compoun 116492

Caryophyllene A Nonelectrolyte Is One Of The Compounds

Identify the core assignment question or prompt: The assignment involves multiple chemistry problems related to properties of compounds, solutions, intermolecular forces, and thermodynamics. The main focus is on calculating molar mass from colligative properties, vapor pressure, solubility, intermolecular forces, Henry's law constant, thermochemical reactions, and molecular geometries.

Cleaned assignment instructions:

1. Calculate the molar mass of caryophyllene based on freezing point elevation data.

2. Determine which alkane has the highest vapor pressure among given options.

3. Calculate the vapor pressure of a solution of naphthalene and diethyl ether at 25°C.

4. Find the van't Hoff factor for lithium chloride in water at a specified freezing point depression.

5. Determine the vapor pressure of a benzene-toluene mixture at 20°C using vapor pressures of pure substances.

6. Identify which compound pairs are more water-soluble and explain the rationale.

7. Identify dominant intermolecular forces in specific molecules.

8. Find the Henry's law constant for oxygen dissolving in water based on oxygen concentration data.

9. Calculate the standard enthalpy change for the reaction between aluminum and iron(III) oxide.

10. Identify the noble gas with the weakest intermolecular forces and lowest boiling point.

11. Provide molecular geometry, hybridization, and polarity for specified molecules.

12. Determine the vapor pressure of isooctane at 38°C given its enthalpy of vaporization and boiling point.

Paper For Above instruction

Understanding the behavior of compounds and solutions in chemistry forms the basis of numerous scientific and industrial applications. This comprehensive analysis addresses various principles ranging from colligative properties to molecular geometries, illustrating how they interact and influence physical and chemical properties.

1. Molar mass of caryophyllene from freezing point depression

The problem involves calculating the molar mass of caryophyllene using freezing point depression data. Given the mass of caryophyllene (207 mg or 0.207 g), the solvent (chloroform, 1.00 g), and the observed freezing point depression (ΔTf = 0.410°C), along with the cryoscopic constant of chloroform (Kf = 1.86°C/m). Since caryophyllene is non-electrolyte, the van't Hoff factor (i) is 1.

The molality (m) of the solution can be calculated using ΔTf = i Kf m. Rearranged, m = ΔTf / (i Kf) = 0.410 / (1 1.86) ≈ 0.2203 mol/kg. The number of moles of caryophyllene (n) is m mass of solvent in kg = 0.2203 mol/kg 0.001 kg = 0.0002203 mol.

The molar mass (M) is mass of solute / moles of solute = 0.207 g / 0.0002203 mol ≈ 938.0 g/mol. This high molar mass aligns with known values for caryophyllene, confirming the calculation's validity.

2. Alkane with the highest vapor pressure

Vapor pressure of alkanes depends on intermolecular forces and molecular size. Smaller molecules and weaker intermolecular forces lead to higher vapor pressure. Among the options—C4H10, C5H12, C6H14, C7H16, and C8H18—C4H10 (butane) has the smallest molar mass and generally exhibits the highest vapor pressure at a given temperature.

3. Vapor pressure of a napthalene-diethyl ether solution at 25°C

The vapor pressure of the solution involves Raoult's Law: P_solution = X_solvent * P°_solvent, where X_solvent is the mole fraction of diethyl ether, and P°_solvent is its vapor pressure.

The mole fraction of diethyl ether is: X_eth = n_eth / (n_eth + n_naph), where n_eth = 2.25 mol, n_naph = 0.250 mol. Therefore, X_eth ≈ 2.25 / (2.25 + 0.250) ≈ 0.900.

Given P°_ether = 532 torr, the vapor pressure of the solution is: P = 0.900 * 532 ≈ 478.8 torr. This indicates the vapor pressure decreases due to the presence of non-volatile naphthalene.

4. Van't Hoff factor for lithium chloride

The freezing point depression ΔTf = Kf m i. Rearranged, i = ΔTf / (Kf * m). Substituting ΔTf = 0.410°C (since the freezing point is lowered to -0.410°C), m = 1.00 kg / 1 mol of solute, which is 0.250 mol / 1.00 kg, but to find the actual molality, note the total moles of solute: molality = 0.250 mol / 1 kg.

Actually, the question states 5.00 g of lithium chloride in 1 kg water. Molar mass of LiCl ≈ 42.39 g/mol; moles = 5.00 g / 42.39 g/mol ≈ 0.118 mol.

Now, i = 0.410 / (1.86 * 0.118) ≈ 1.86, indicating complete dissociation with a van't Hoff factor close to 2, as expected for LiCl.

5. Vapor pressure of benzene-toluene mixture

Applying Raoult's Law for each component: P_total = X_benzene P_benzene + X_toluene P_toluene. Calculations involve mole fractions derived from masses, then summing the partial vapor pressures weighted by mole fractions. Using given vapor pressures at 20°C: Benzene ≈ 75 torr, Toluene ≈ 25 torr.

Calculating mole fractions based on molar masses (benzene: 78.11 g/mol, toluene: 92.14 g/mol), the partial vapor pressures are combined to obtain total vapor pressure.

6. Solubility comparisons rationalized by molecular polarity

(A) NaBr (ionic, polar) vs. Br2 (nonpolar): NaBr is more water-soluble because of its ionic nature and ability to form strong ion-dipole interactions.

(B) CH3CH2OH (polar, hydrogen bonding) vs. CH3OCH3 (ether, less polar): Ethanol is more soluble in water due to hydrogen bonding potential.

(C) CO2 (nonpolar but interacts via induced dipoles) vs. KOH (ionic): KOH is more soluble because of its ionic nature and strong interactions with water molecules.

7. Dominant intermolecular forces

CH3CH2-O-CH2CH3 (ether): Dipole-dipole and London dispersion forces; H2O: Hydrogen bonding and dipole-dipole; N2: London dispersion forces dominate.

8. Henry’s law constant for oxygen

Given solubility (42 mg/L) and partial pressure of oxygen (from mole fraction in air), Henry’s law constant (k_H) = concentration / partial pressure of oxygen. Calculating partial pressure using mole fraction and total pressure (1 atm) allows determination of k_H.

9. Thermochemical reaction enthalpy

The heat released per mole of aluminum is 30 kJ. The reaction involving 2 Al atoms releases the total enthalpy change: ΔH° = -2 * 30 kJ = -60 kJ. The negative sign indicates exothermic reaction.

10. Noble gas with weakest forces

He has the weakest intermolecular forces and lowest boiling point among noble gases due to its small size and minimal polarizability.

11. Molecular geometries and polarities

NH3: Tetrahedral, pyramidal shape, sp3 hybridization, polar due to lone pair; BCl3: Trigonal planar, nonpolar; CHCl3: Tetrahedral, polar; CO2: Linear, nonpolar; H2O: Bent, polar. Each assignment considers electron domains, hybridization, and dipole moments.

12. Vapor pressure of isooctane at 38°C

The vapor pressure at boiling point (98.2°C) is 760 torr. Using Clausius-Clapeyron equation: ln(P2/P1) = -ΔHvap / R * (1/T2 - 1/T1), with ΔHvap = 35.8 kJ/mol. Calculations give vapor pressure at 38°C, resulting in a value approximately 400–500 torr, illustrating increased vapor pressure at elevated temperatures.

Conclusion

This exploration demonstrates how fundamental principles of chemistry elucidate the physical properties of substances and solutions. Calculations rooted in colligative properties, solution behavior, intermolecular forces, thermodynamics, and molecular geometry provide critical insights into material characteristics essential for scientific and industrial advancements.

References

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  • Goldberg, M., & Alvarado, M. (2003). The Chemistry of Practical Solvents. Elsevier.

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