CHEM 1115 Take-Home Test 1 Study Guide

CHEM 1115 Take-Home Test 1 Notes

General Information

Due Date: Wednesday, February 15 by 11 AM
Name:

Useful Information

Solubility: solubility = k imes P
Molarity relation: m = ext{moles/kg}
Ideal Gas Law: PV = nRT
Concentration relation: c = rac{n}{V}
Distance relation for a triangle: c^2 = a^2 + b^2
Show work for credit.

Problem-Solving Sections

Problem 1 (3 points)

Question: What types of solvents (polar, nonpolar, or none) should dissolve the following compounds?
(a) CH3OH

Solubility Type: Polar solvent;
Reason: Contains a hydroxyl group (–OH) which allows hydrogen bonding.
(b) PCl3
Solubility Type: Polar solvent;
Reason: Has a trigonal pyramidal shape, leading to a dipole moment.
(c) Na2SO4
Solubility Type: Polar solvent;
Reason: Ionic compound, dissolves in water due to strong ion-dipole interactions.

Problem 2 (6 points)

Use the phase diagram above to answer the following questions:
(a) At point A (5,000 psi and -75 °C), what phase(s) are present?

Answer: Determine from the phase diagram, likely solid and gas or mixed phase.
(b) What is the triple point of this compound?
Answer: Identify coordinate point where all three phases coexist from the phase diagram.
(c) If you change the conditions from those at point A to those at point C (-25 °C, 20,000 psi)?
Answer: Determine the phase based on changes in pressure and temperature.

Problem 3 (4 points)

Question: Sodium has a radius of 185 pm and a simple cubic unit cell. What is the density of sodium?**

Formula for density: ext{Density} = rac{ ext{mass of unit cell}}{ ext{volume of unit cell}}
Unit cell volume for simple cubic: a^3 , where a = 2 imes radius = 2 * 185 pm = 370 pm = 3.7 imes 10^{-8} cm
Density Calculation Steps:
1. Calculate volume: (3.7 imes 10^{-8})^3 ext{ cm}^3
2. Mass of unit cell: Calculate using the molar mass and Avogadro's number.
3. Final density calculation.

Problem 4 (2 points)

Question: If a brittle solid is soluble in water, but not nonpolar solvent, has a high melting point, and is white, what type of solid is it and why?**

Type of Solid: Ionic solid.
Reason: Has strong ionic bonds, soluble in polar water due to ion-dipole interactions, high melting point due to lattice energy, and typically appears white due to its crystal structure.

Problem 5 (3 points)

Draw the dipole moment for the following species. If it has no dipole moment, write ‘nonpolar’.
(a) PF4-

Dipole Moment: Draw structure; polar due to charge distribution.
(b) NO2-
Dipole Moment: Draw structure; polar due to bent geometry.
(c) BeF2
Dipole Moment: Write ‘nonpolar’; linear structure cancels dipoles.

Problem 6

(3 points)
Draw a π bonding orbital. Explain what makes it a pi bond.

Representation: Draw orbital; shape resembles two lobes above and below the bonding axis.
Explanation: Formed by the side-to-side overlap of p orbitals. Pi bonds are weaker than sigma bonds and occur alongside them in double or triple bonds.

Molecular Orbital Theory Section

Use the molecular orbital diagram shown to answer questions 7-9.

Atomic orbitals: Refer to the provided diagram.
Molecular orbitals: Identify bonding and antibonding orbitals.

Problem 7**

Question: What is the bond order for O2, F2, and Ne2?**

Bond Order Calculation Formula: ext{Bond Order} = rac{ ext{(number of bonding electrons - number of antibonding electrons)}}{2}
- O2: Approximately 2
- F2: Approximately 1
- Ne2: 0 (molecular orbital does not exist)

Problem 8**

Which, if any, of these molecules is paramagnetic and why?

Paramagnetic Molecule(s): O2
- Reason: Contains unpaired electrons in molecular orbitals, leading to magnetic behavior.
- F2 and Ne2: All electrons are paired, thus diamagnetic.

Problem 9

Write out the molecular orbital description for O2, F2, and Ne2.
(a) O2:

Electronic Configuration: (σ{2s})^2 (σ^{2s})^2 (σ{2pz})^2 (π{2px})^2 (π{2py})^2 (π^{2px})^1 (π^*{2py})^1
Bond Order: 2
(b) F2:
Electronic Configuration: (σ{2s})^2 (σ^{2s})^2 (σ{2pz})^2 (π{2px})^2 (π{2py})^2 (π^{2px})^2 (π^*{2py})^0
Bond Order: 1
(c) Ne2:
Electronic Configuration: All electrons paired, leading to bond order = 0.

Problem 10 (4 points)

During an ice storm, salt is used to keep the water from freezing on the road. If K_f = 1.853 \text{ºC/m} for water, how much salt is required to lower the freezing point of 10,000 kg of water to -3 ºC?

Freezing Point Depression Formula: \Delta Tf = Kf imes m imes i
- Where: Kf is freezing point depression constant, m is molality, and i is van 't Hoff factor (for salt, i = 2 for NaCl).
- Calculation Steps:
1. \Delta Tf = Tf^0 - T_f = 0 - (-3) = 3
2. Rearrange to solve for m: m = \frac{\Delta Tf}{Kf \times i}
3. Compute mass of salt needed with m and water mass.

Problem 11 (4 pts)

The vapor pressure of water is 1074 mm of Hg at 110 °C. What concentration of sucrose (C12H22O11, MW=342 g/mole) would be required to make the normal boiling point of the solution 110 °C?

Boiling Point Elevation Formula: \Delta Tb = Kb imes m imes i
- Where: Kb is boiling point elevation constant, determine based on water properties.
- Calculation Steps:
1. \Delta Tb = Tb - T_b^0 = 110 - 100 = 10
2. Rearrange to solve for m as shown above.
3. Find mass of sucrose needed and convert to desired units.

Problem 12 (3 pts)

Convert a 0.50 mole fraction solution of methanol (CH3OH) in water to molality.

Definition of Molality: m = \frac{\text{moles of solute}}{\text{kg of solvent}}
Steps to Calculate:
1. Use the mole fraction: X{solvent} + X{solute} = 1$$
2. Calculate moles of each component.
3. Solve for molality using the found values.

To interpret the phase diagram effectively, follow these steps:

Identify Axes: Determine what the x-axis and y-axis represent, typically temperature and pressure.
Locate Points and Regions: Find specific points (such as point A) and observe the different regions (solid, liquid, gas) separated by the lines (phase boundaries).
Understand Phase Changes: Recognize that moving across different regions indicates a phase change (e.g., solid to liquid, liquid to gas).
Triple Point: Identify the triple point where all three phases coexist.
Critical Point: Understand the significance of the critical point, above which distinct liquid and gas phases do not exist.