Exam1 SP22 practice key (1)

Chem121 General Chemistry II Spring 2022
Exam 1 (100 points)
8 February 2022
Name_______________________________________________________
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Useful information for this exam
CONSTANTS
R = 8.3145 J·mol-1·K-1 = 0.08206 L·atm·mol-1·K-1 = 0.08206 atm·M-1·K-1
NA = 6.022 × 1023 mol-1
h = 6.626 × 10-34 J·S
c = 2.998 × 108 m·s-1
֯C = K – 273.15
1 atm = 760 torr = 760 mmHg = 14.7 psi = 1.01325bar
1cal = 4.184J
Partial credit will be given where possible and appropriate, so show calculations, and include
units and sig figs.
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1. (4 points) Arrange these compounds in order of increasing boiling point.
a. CH4
b .CH3CH3
c. CH3CH2OH
Lower b.p. (
a < b< c
) < (
) < (
) Higher b.p.
2. (5 points) Which compound would you expect to have greater surface tension: acetone
(CH3COCH3) or water (H2O)? What is the name of the intermolecular force that makes
the difference between these two? (the force that exists in one compound but not in the
other one)
Water
Hydrogen bonding
3. (6 points) Determine whether each pair of compounds forms a homogeneous solution
when combined. (Choose between Homogeneous solution and No mixing)
a) CCl4 and H2O (Homogeneous solution / No mixing)
b) Br2 and CH3CH2CH2CH3 (Homogeneous solution / No mixing)
c) CH3OH and CH3CH2CH2CH2CH3 (Homogeneous solution / No mixing)
4. (9 points) Dr. Zoidberg made a 0.732 M Glucose (aq) solution at 25˚C. What is the
molality of this solution? Molecular weight of Glucose is 180.2 g/mol and density of the
solution at 25˚C is 1.12 g/mL.
0.732 M means 0.732 mol glucose / 1 L solution.
1L solution x (1000mL/1L) x (1.12 g/mL) = 1120 g solution
within the solution, we have 0.732 mol solute by definition which is
0.732 mol solute x (180.2 g/mol) = 131.91 g solute
g solvent = g solution – g solute = 1120g – 131.91 g = 988 g
Molality = mol solute / kg solvent = 0.732 mol / 0.988 kg = 0.74 m
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5. (5 points) When lithium iodide (LiI) is dissolved in water, the solution becomes hotter.
a) Is the dissolution of lithium iodide endothermic or exothermic?
a. endothermic
b. exothermic
b) If we describe the solution formation enthalpy as ΔHsln = ΔHsolute + ΔHhydration,
which of the following statement is true?
a. ΔHsolute < 0
b. ΔHhydration < 0
c. ΔHsln > 0
d. The energy is being absorbed by the solution.
6. (14 points) A solution contains 45.0%mass of heptane (C7H16) and 55.0 %mass of
octane (C8H18) at 25˚C. The vapor pressures of pure heptane and pure octane at 25˚C are
45.8 torr and 10.9 torr, respectively.
a) Assuming the ideal behavior, what is the total pressure above the solution? 9 points
Assuming 100 g total solution,
Mol heptane = 45.0 g x (1mol/100.21g) = 0.4491 mol
Mol octane = 55.0 g x (1mol/114.23g) = 0.4815 mol
Mol ratio heptane = 0.4491 mol / (0.4491 mol + 0.4815 mol) = 0.4826
Mol ratio octane = 1 – mol ratio heptane = 1-0.2753 = 0.5174
Vapor pressure heptane = 0.4826 x 45.8 torr = 22.10 torr
Vapor pressure octane = 0.5174 x 10.9 torr = 5.640 torr
Total vapor pressure = 22.10 torr + 5.640 torr = 27.7 torr
b) What is the mole fraction of heptane in the vapor?
Mol fration = pressure fraction in gas
22.10 torr / (22.10 torr + 5.640 torr) = 0.797
[To be fair, this problem was supposed to be a challenging problem that only those who
tried home work diligently can do]
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7. (3 points) What can you do to dissolve more carbon dioxide gas (CO2) into water?
a. Increase the pressure of CO2
b. Decrease the pressure of CO2
c. Increase the temperature
d. None of the above
8. (9 points) Calculate the vapor pressure of a solution containing 24.5 g of glycerine
(C3H8O3, molar mass – 92.09 g/mol) in 2.00 x 102 mL of water (H2O, molar mass – 18.01
g/mol) at 62.0˚C. The vapor pressure of pure water at this temperature is 33.8 torr.
Assume that glycerin is nonvolatile and dissolves molecularly. Density of water at 62.0˚C
is 0.980 g/mL.
Mol glycerine = 24.5 g x (1mol/92.09g) = 0.2660 mol
Mol water = 2.00 x 102 mL x (0.980 g / mL) x (1mol/18.01g)= 10.88 mol
Mol fraction water = 10.88 mol / (0.2660 mol + 10.88 mol) = 0.9761
Vapor pressure = 0.9761 x 33.8 torr = 33.0 torr
9. (5 points) You took 10.0 mL of 233 mM HCl (aq) and added water to make it 80.0 mL
final. What is the molarity of the diluted solution in mM?
M1V1 = M2V2
233 mM x 10.0 mL = M2 x 80.0mL
M2 = 29.1 mM
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10. (9 points) What mass of salt (NaCl) should you add to 1.00 L of water in an ice-cream
maker to make a solution that freezes at = -10.0˚C? Assume complete dissociation of the
NaCl and density of 1.00 g/mL for water.
Van’t Hof factor i for NaCl is 2.
ΔTf = 10.0°C
ΔTf = 2 x m x Kf
m = 10.0°C / (2 x 1.86 °C/m) = 2.688 m
mol NaCl = 2.688 m x (1.00 L x 1.00 g/mL) = 2.688 mol
g NaCl = 2.688 mol x [(22.99 + 35.45) g / mol]
= 157 g
11. (6 points) List the following solutions from the highest to lowest boiling points.
a. 0.50 m C12H22O11 (aq)
b. 0.50 m MgSO4 (aq)
c. 0.50 m K2PO4 (aq)
d. Pure water, H2O
Highest b.p. (
C > b > a> d
)>(
)>(
)>(
) > Lowest b.p.
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12. (9 points) Osmotic pressure of an unknown protein solution containing 2.11 mg of an
unkown protein per 10.0 mL of solution is 2.45 torr at 25˚C. Find the molar mass of the
unkown protein.
P = 2.45 torr x (1atm/760torr) = 0.003224 atm [1]
Pi = MRT
M = Pi/RT = (0.003224 atm) / (0.08206 L·atm·mol-1·K-1 x (273.15+25 K)) = 0.000132 M
Mol protein = 0.000132 M x 10.0 mL x (1L/1000mL) = 1.3176 x 10-6 mol
[1 for knowing molar mass calc]
Molar mass = 2.11 mg x (1g/1000mg) / 1.3176 x 10-6 mol = 1.60 x 103 g/mol
13. (6 points) Consider the reaction:
8H2S(g) + 4 O2(g) → 8H2O(g) + S8(g)
If Δ[H2S]/Δt = - 0.080 M/s what is the Δ[O2]/Δt ?
Rate = - (1/8) x (- 0.080 M/s) = 0.010 M/s
Δ[O2]/Δt = - 4 x rate = - 0.040 M/s
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14. (10 points) Consider the following reaction.
a) By reading the graph, calculate the average rate between time 0 and 20 seconds.
Δ[H2O2]/Δt =(0.55 M – 1.00 M) / (20s – 0 s) = -0.023 M/s
Rate = -(1/2) x Δ[H2O2]/Δt = 0.012 M/s
b) Choose the correct statement about instantaneous d[H2O2]/dt in the graph.
a. Instantaneous rate of H2O2 consumption at t=20s is faster than t=70s.
b. Instantaneous rate of H2O2 consumption at t=20s is slower than t=70s.
c. Instantaneous rate of H2O2 consumption at t=20s is about the same as t=70s.
d. Instantaneous rate of H2O2 consumption is constant in this experiment.
[Be able to make an approximate parallel lines for slopes to compare]
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