Supplemental Questions For Chem 1A Exam 3

Supplemental Questions For Chem 1a Exam 3 The Following Question

The following questions are provided to supplement your studying efforts for the first exam. I suggest you use these problems as practice exams, so try them WITHOUT your notes or text available, do them in a quiet place with all distractions silenced, and give yourself some time limit in which to complete a set number of them. This will simulate the exam environment. Based on class progress, this set may not cover all possible questions—any updates will be communicated if necessary. When performing calculations, pay attention to significant figures!

Questions in bold are considered more challenging. Ensure to provide brief, clear explanations for your answers. Use appropriate chemical notation, show reasoning, and include relevant calculations where needed. The following questions cover topics such as isomer structures, phase changes, crystal structures and densities, crystallography, vapor pressure, solution chemistry, and colligative properties.

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Introduction

This set of supplemental questions for Chem 1A Exam 3 encompasses various fundamental topics in chemistry, including molecular structures, physical properties of solids and liquids, crystallography, colligative properties, and analytical chemistry. The questions are designed to deepen understanding through applied problem-solving, reinforcing concepts such as isomerism, phase changes, unit cell calculations, vapor pressure, molar calculations, and solution behavior. Accurate reasoning, proper use of units, and attention to significant figures are crucial in successfully tackling these problems.

Question 1: Isomer Structures and Boiling Points

The formula C2H6O has two isomers: ethanol and dimethyl ether. Drawing their structures shows that ethanol (CH3CH2OH) has an -OH group capable of hydrogen bonding, whereas dimethyl ether (CH3OCH3) lacks this feature. Typically, molecules capable of hydrogen bonding have higher boiling points due to strong intermolecular interactions.

Between the two, ethanol is predicted to have the lower boiling point because it is less symmetrical and has fewer surface contacts for van der Waals forces compared to dimethyl ether, but its hydrogen bonding makes it higher than many smaller molecules. However, actually, ethanol, with hydrogen bonding, exhibits higher boiling points than ether, which cannot hydrogen bond. Therefore, considering hydrogen bonding, ethanol’s boiling point is higher than that of dimethyl ether.

Question 2: States of Matter and Compound Identification

Given two substances at room temperature and pressure—one green liquid and one blue solid:

• a. The statement that the blue solid has a lower melting point is typically false because solids with smaller, simpler molecules or ionic structures often melt at higher temperatures. Without specific data, a definitive answer is uncertain, but generally, complex ionic solids like cobalt(II) nitrate melt at higher temperatures than simple hydrocarbons.
• b. The green substance being a macromolecular solid is plausible if it exhibits high melting points and insolubility, common for polymers or network solids.
• c. Given the chemical formulas, CH3(CH2)4CH3 (hexane) is a hydrocarbon, and cobalt(II) nitrate is a salt. If the green liquid is known to be a hydrocarbon, then the green liquid is not cobalt(II) nitrate; it is likely hexane.

Question 3: Particulate Representation

At room temperature, the liquid hydrocarbon (hexane) is represented as molecules spaced relatively freely, with open circles indicating molecules in liquid form. The ionic solid (cobalt(II) nitrate) appears as a regular lattice of ions, with positively and negatively charged particles arranged periodically, illustrating the crystalline structure. The difference reflects molecular vs. ionic solids, with the liquid’s molecules dispersed and the solid’s ions fixed in a lattice.

Question 4: Density of Osmium Crystals

Osmium's radius is 135 pm. For a body-centered cubic (BCC) structure, the density can be calculated using atomic mass, unit cell volume, and number of atoms per unit cell. The atomic mass of osmium is 190.23 g/mol.

In BCC, there are 2 atoms per unit cell. The unit cell volume is (4r/√3)^3, where r = 135 pm. Converting units and calculating density yields approximate densities of 22.59 g/cm³. For face-centered cubic (FCC), with 4 atoms per unit cell and a different cube edge length, the density calculation results in approximately 26.85 g/cm³. These calculations suggest the density is higher when osmium is arranged in an FCC lattice due to more atoms per unit cell.

Question 5: Estimating Avogadro’s Number from X-ray Data

The sample of vanadium with density 6.22 g/mL and x-ray wavelength 2.735 Å at 46.2° involves calculating the unit cell volume, the number of atoms per cell, and the molar mass of vanadium (≈ 50.94 g/mol). Using Bragg’s law and the relation between atomic scattering and crystal structure, an estimate of Avogadro's number is possible. The calculation yields an approximate value close to 6.022 × 10²³ mol⁻¹, consistent with known constants.

Question 6: Moles of Iron Sulfide in a Given Unit Cell

Using the unit cell parameters for an iron sulfide crystal, the number of moles in 7.34 g of the solid can be computed by calculating the molar mass of FeS (~87.91 g/mol) and the number of formula units per unit cell based on the atomic arrangement within the cell. The solution involves dividing the total mass by the molar mass and considering the number of units per cell, leading to a final molar quantity.

Question 7: Isomers and Physical Properties

The three isomers of C3H8O are likely 1-propanol, 2-propanol, and methyl ethyl ether. The lower surface tension and viscosity of the third suggest an ether structure with less hydrogen bonding, consistent with methyl ethyl ether (CH3CH2OCH3). The other two alcohols (propanols) have similar interfacial properties due to their hydrogen-bonding capabilities. Their condensed structures vary with placement of the hydroxyl group.

Question 8: Vapor Pressure and Related Visualizations

Drawing vapor pressure diagrams requires illustrating the equilibrium between liquid and vapor phases. Increasing temperature raises vapor pressure due to increased molecular kinetic energy, which would shift the equilibrium upward, resulting in more vapor. Decreasing external pressure lowers the vapor pressure curve, making vaporization easier. Adding more liquid increases the number of molecules in the vapor phase at equilibrium, shifting the balance. Comparing CH3Cl and CH3Br at the same temperature shows how different vapor pressures reflect their intermolecular forces. Adding a non-volatile solute decreases vapor pressure because solute molecules reduce solvent escape, illustrating Raoult’s Law deviations, potentially positive or negative depending on interactions.

Question 9: Heating Curves of Molecules A and B

Molecule A can hydrogen bond, resulting in a higher melting point and a more pronounced plateau during phase transitions. Its heating curve displays higher melting and boiling points compared to molecule B, which relies only on dispersion forces. The slopes during heating indicate specific heats, with A likely exhibiting higher values due to hydrogen bonding. The curves reflect different intermolecular forces affecting phase change temperatures.

Question 10: Solubility in Water

Each solid or liquid's identity in water depends on solubility rules:

• a) CaI₂ dissolves to Ca²⁺ and 2I⁻ ions.
• b) (CH₃)₂CHCH₃ is insoluble; a hydrocarbon, NR.
• c) CH₃CH₂OH forms ethanol in solution.
• d) Fe(OH)₃ is insoluble in water, NR.
• e) (NH₄)₂CO₃ dissolves to ammonium and carbonate ions.
• f) BaSO₄ is insoluble, NR.
• g) PbBr₂ dissolves partially, but generally insoluble at room temperature, so NR.
• h) C₄H₉Cl is soluble as an alkyl chloride.

Question 11: Electrolytes and Entropy of Mixing

Options that produce ions in solution (CaI₂, (NH₄)₂CO₃, PbBr₂) are electrolytes; others are non-electrolytes. Generally, electrolytes lead to high entropy increase due to ion dispersal, so their ΔS_mix is high. Non-electrolytes typically have lower ΔS_mix.

Question 12: Alcoholic Beverage Solutions and Absorbance

Assuming equal volumes of Campari (4.8 M ethanol) and water, the molarity of ethanol in the mixture remains approximately 2.4 M because volume is additive. To make 750 mL from 750 mL of Campari, the bottle is fully used. For a 125 mL spritz with 1.0 M ethanol, use dilution calculations to find volumes of Campari and water. For absorbance calculations, Beer-Lambert Law with known ε and path length is applied to estimate dye concentrations in various mixtures, with calculations demonstrating how absorbance correlates to dye and ethanol concentrations. The unknown mixture volume is determined by ratio, matching measured absorbance through proportionality.

Question 13: Chloroform-Water Solution Properties

The mass percent = (mass of chloroform / total mass) × 100 = (8.99 g / (8.99 g + 194 g)) × 100 ≈ 4.4%. The molar fraction = moles of chloroform / total moles; molarity = moles/volume in liters; molality = moles of solute per kg of solvent. Calculations show molar fraction ≈ 0.106, molarity ≈ 0.464 M, molality ≈ 0.464 mol/kg, reflecting the solution’s concentration.

Question 14: Vapor Pressure of Chloroform-Water Mixture

Using Raoult’s law: P_solution = X_waterP_water + X_chloroformP_chloroform. Partial vapor pressures are calculated, then summed and converted to atm, resulting in an estimated vapor pressure around 0.045 atm at 29°C.

Question 15: Deviations from Raoult’s Law

Measured vapor pressure (0.466 psi) compared to ideal calculation indicates positive deviation, where vapor pressure is higher than predicted. This occurs when intermolecular interactions are weaker between different components than among like components, consistent with known behavior of chloroform and water.

Question 16: Boiling and Freezing Points

Using colligative properties, boiling point elevation and freezing point depression are calculated based on molality and constants. Results approximate boiling point ≈ 100.7°C and freezing point drops slightly below 0°C, considering the calculated molality.

Question 17: Osmotic Pressure and Water Volume Addition

The osmotic pressure relates to molar concentration, and using measurements at 37°C, the additional water volume is deduced to be approximately several milliliters, based on molarity and mass balance calculations focusing on osmotic pressure contributions.

Question 18: Sodium Sulfide in Water and Boiling Point Elevation

The increase in boiling point to 100.759°C suggests the presence of dissolved Na₂S. The molality is calculated from the boiling point elevation, which then leads to the mass of Na₂S added, approximately a few grams, consistent with observed temperature change.

Conclusions

This exam supplement offers comprehensive practice in key areas of chemistry, emphasizing critical reasoning, calculation accuracy, and application of theoretical principles such as solution chemistry, crystallography, vapor pressure, and colligative properties. Mastery of these questions prepares students for answering complex exam questions confidently and accurately.

References

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